RS Aggarwal Solutions for Class 11 Chapter 15 Trigonometric, Or Circular, Functions

Exercise 15(A)

Question 1. If cos θ = -√3/2 and θ lies in Quadrant III, find the value of all the other five trigonometric functions.
Answer: Given: \( \cos \theta = -\frac{\sqrt{3}}{2} \) Since θ is in the IIIrd Quadrant, both sin and cos will be negative, while tan will be positive. We apply the fundamental identity: \( \cos^2 \theta + \sin^2 \theta = 1 \) Substituting the given value: \( \left( -\frac{\sqrt{3}}{2} \right)^2 + \sin^2 \theta = 1 \) \[ \implies \frac{3}{4} + \sin^2 \theta = 1 \] \[ \implies \sin^2 \theta = 1 - \frac{3}{4} \] \[ \implies \sin^2 \theta = \frac{4-3}{4} \] \[ \implies \sin^2 \theta = \frac{1}{4} \] \[ \implies \sin \theta = \pm \frac{1}{2} \] Since θ lies in the IIIrd quadrant where sine is negative: \[ \therefore \sin \theta = -\frac{1}{2} \] Now, using the definition of tangent: \( \tan \theta = \frac{\sin \theta}{\cos\theta} \) Substituting values: \( \tan \theta = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \) \[ = -\frac{1}{2} \times \left( -\frac{2}{\sqrt{3}} \right) \] \[ = \frac{1}{\sqrt{3}} \] For cosecant: \( \cosec \theta = \frac{1}{\sin \theta} \) Substituting values: \( \cosec \theta = \frac{1}{-\frac{1}{2}} \) \[ = -2 \] For secant: \( \sec \theta = \frac{1}{\cos\theta} \) Substituting values: \( \sec \theta = \frac{1}{-\frac{\sqrt{3}}{2}} \) \[ = -\frac{2}{\sqrt{3}} \] For cotangent: \( \cot \theta = \frac{1}{\tan \theta} \) Substituting values: \( \cot \theta = \frac{1}{\frac{1}{\sqrt{3}}} \) \[ = \sqrt{3} \] Hence, the values of other trigonometric functions are:

In simple words: When one trigonometric function is given and you know which quadrant the angle is in, use the basic identity to find another function. Then use ratios and reciprocal relations to calculate all remaining functions.

Exam Tip: Always check the quadrant first - it tells you the sign of each function. Use the Pythagorean identity as your starting point for finding missing values.

Question 2. If sin θ = -1/2 and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Answer: Given: \( \sin \theta = -\frac{1}{2} \) Since θ is in the IVth Quadrant, sine and tan will be negative but cos will be positive. We apply the fundamental identity: \( \sin^2 \theta + \cos^2 \theta = 1 \) Substituting the given value: \( \left( -\frac{1}{2} \right)^2 + \cos^2 \theta = 1 \) \[ \implies \frac{1}{4} + \cos^2 \theta = 1 \] \[ \implies \cos^2 \theta = 1 - \frac{1}{4} \] \[ \implies \cos^2 \theta = \frac{4-1}{4} \] \[ \implies \cos^2 \theta = \frac{3}{4} \] \[ \implies \cos \theta = \pm \frac{\sqrt{3}}{2} \] Since θ lies in the IVth quadrant where cosine is positive: \[ \therefore \cos \theta = \frac{\sqrt{3}}{2} \] Now, using the definition of tangent: \( \tan \theta = \frac{\sin \theta}{\cos\theta} \) Substituting values: \( \tan \theta = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \) \[ = -\frac{1}{2} \times \left( \frac{2}{\sqrt{3}} \right) \] \[ = -\frac{1}{\sqrt{3}} \] For cosecant: \( \cosec \theta = \frac{1}{\sin \theta} \) Substituting values: \( \cosec \theta = \frac{1}{-\frac{1}{2}} \) \[ = -2 \] For secant: \( \sec \theta = \frac{1}{\cos\theta} \) Substituting values: \( \sec \theta = \frac{1}{\frac{\sqrt{3}}{2}} \) \[ = \frac{2}{\sqrt{3}} \] For cotangent: \( \cot \theta = \frac{1}{\tan \theta} \) Substituting values: \( \cot \theta = \frac{1}{-\frac{1}{\sqrt{3}}} \) \[ = -\sqrt{3} \] Hence, the values of other trigonometric functions are:

In simple words: Start by using the Pythagorean identity to find cosine from the given sine value. Remember that your quadrant determines whether each ratio is positive or negative. Then apply reciprocal and ratio formulas to find the remaining functions.

Exam Tip: Always verify that your final answer respects the quadrant rule - in Quadrant IV, only cos and sec should be positive. Double-check reciprocal pairs multiply to 1.

Question 3. If cosec θ = 5/3 and θ lies in Quadrant II, find the values of all the other five trigonometric functions.
Answer: Given: \( \cosec \theta = \frac{5}{3} \) Now, we understand that: \( \sin \theta = \frac{1}{\cosec\theta} \) Substituting values: \( \sin \theta = \frac{1}{\frac{5}{3}} \) \[ \sin \theta = \frac{3}{5} \quad \text{...(i)} \] We know that: \( \sin^2 \theta + \cos^2 \theta = 1 \) Substituting the values: \( \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \) \[ \implies \frac{9}{25} + \cos^2 \theta = 1 \] \[ \implies \cos^2 \theta = 1 - \frac{9}{25} \] \[ \implies \cos^2 \theta = \frac{25-9}{25} \] \[ \implies \cos^2 \theta = \frac{16}{25} \] \[ \implies \cos \theta = \pm \frac{4}{5} \] Since θ lies in the IInd quadrant where cosine is negative: \[ \therefore \cos \theta = -\frac{4}{5} \] Now, using the definition of tangent: \( \tan \theta = \frac{\sin \theta}{\cos\theta} \) Substituting values: \( \tan \theta = \frac{\frac{3}{5}}{-\frac{4}{5}} \) \[ = \frac{3}{5} \times \left( -\frac{5}{4} \right) \] \[ = -\frac{3}{4} \] For secant: \( \sec \theta = \frac{1}{\cos\theta} \) Substituting values: \( \sec \theta = \frac{1}{-\frac{4}{5}} \) \[ = -\frac{5}{4} \] For cotangent: \( \cot \theta = \frac{1}{\tan \theta} \) Substituting values: \( \cot \theta = \frac{1}{-\frac{3}{4}} \) \[ = -\frac{4}{3} \] Hence, the values of other trigonometric functions are:

In simple words: When you are given a reciprocal function like cosecant, take its reciprocal to get sine. Then apply the Pythagorean identity to find cosine, keeping track of the quadrant's sign rule. Finally, use ratios to find the remaining functions.

Exam Tip: When given a reciprocal function (cosec, sec, or cot), always convert it first to get the primary function (sin, cos, or tan). This makes the rest of the problem straightforward.

Question 4. If sec θ = √2 and θ lies in Quadrant IV, find the values of all the other five trigonometric functions.
Answer: Given: \( \sec \theta = \sqrt{2} \) Since θ is in the IVth Quadrant, sine and tan will be negative but cos will be positive. Now, we understand that: \( \cos \theta = \frac{1}{\sec\theta} \) Substituting values: \( \cos \theta = \frac{1}{\sqrt{2}} \quad \text{...(i)} \) We know that: \( \cos^2 \theta + \sin^2 \theta = 1 \) Substituting values: \( \left(\frac{1}{\sqrt{2}}\right)^2 + \sin^2 \theta = 1 \) \[ \implies \frac{1}{2} + \sin^2 \theta = 1 \] \[ \implies \sin^2 \theta = 1 - \frac{1}{2} \] \[ \implies \sin^2 \theta = \frac{2-1}{2} \] \[ \implies \sin^2 \theta = \frac{1}{2} \] \[ \implies \sin \theta = \pm \frac{1}{\sqrt{2}} \] Since θ lies in the IVth quadrant where sine is negative: \[ \therefore \sin \theta = -\frac{1}{\sqrt{2}} \] Now, using the definition of tangent: \( \tan \theta = \frac{\sin \theta}{\cos\theta} \) Substituting values: \( \tan \theta = \frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \) \[ = -\frac{1}{\sqrt{2}} \times (\sqrt{2}) \] \[ = -1 \] For cosecant: \( \cosec \theta = \frac{1}{\sin \theta} \) Substituting values: \( \cosec \theta = \frac{1}{-\frac{1}{\sqrt{2}}} \) \[ = -\sqrt{2} \] For cotangent: \( \cot \theta = \frac{1}{\tan \theta} \) Substituting values: \( \cot \theta = \frac{1}{-1} \) \[ = -1 \] Hence, the values of other trigonometric functions are:

In simple words: Begin by taking the reciprocal of the given secant to find cosine. Use the Pythagorean relation to calculate sine, being careful about the sign based on the quadrant. Then find the remaining functions using ratio and reciprocal definitions.

Exam Tip: Notice that in Quadrant IV, only cosine and secant are positive while sine, tangent, and cotangent are negative. This quadrant rule is essential for getting the signs right.

Question 5. If sin x = -2√6/5 and x lies in Quadrant III, find the values of cos x and cot x.
Answer: Given: \( \sin x = -\frac{2\sqrt{6}}{5} \) To find: cos x and cot x Since x is in the IIIrd Quadrant, both sine and cos will be negative but tan will be positive. We know that: \( \sin^2 x + \cos^2 x = 1 \) Substituting values: \( \left( -\frac{2\sqrt{6}}{5} \right)^2 + \cos^2 x = 1 \) \[ \implies \frac{24}{25} + \cos^2 x = 1 \] \[ \implies \cos^2 x = 1 - \frac{24}{25} \] \[ \implies \cos^2 x = \frac{25-24}{25} \] \[ \implies \cos^2 x = \frac{1}{25} \] \[ \implies \cos x = \pm \frac{1}{5} \] Since x lies in the IIIrd quadrant where cosine is negative: \[ \therefore \cos x = -\frac{1}{5} \] Now, using the definition of tangent: \( \tan x = \frac{\sin x}{\cos x} \) Substituting values: \( \tan x = \frac{-\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} \) \[ = -\frac{2\sqrt{6}}{5} \times (-5) \] \[ = 2\sqrt{6} \] Now, using the definition of cotangent: \( \cot x = \frac{1}{\tan x} \) Substituting values: \( \cot x = \frac{1}{2\sqrt{6}} \) Hence, the values of other trigonometric functions are:

In simple words: Apply the Pythagorean identity to find cosine from the given sine value. Remember that in Quadrant III, cosine is also negative. Then find tangent and cotangent using their respective ratio definitions.

Exam Tip: When squaring a complex radical like \(2\sqrt{6}\), compute \((2\sqrt{6})^2 = 4 \times 6 = 24\) carefully. Always verify your quadrant rules before assigning signs to final answers.

Question 6. If cos x = -√15/4 and π/2 < x < π, find the value of sin x.
Answer: Given: \( \cos x = -\frac{\sqrt{15}}{4} \) To find: value of sin x Given that: \( \frac{\pi}{2} < x < \pi \) So, x lies in IInd quadrant and sin will be positive. We know that: \( \cos^2 \theta + \sin^2 \theta = 1 \) Substituting values: \( \left( -\frac{\sqrt{15}}{4} \right)^2 + \sin^2 \theta = 1 \) \[ \implies \frac{15}{16} + \sin^2 \theta = 1 \] \[ \implies \sin^2 \theta = 1 - \frac{15}{16} \] To calculate this, I first find a common denominator: \[ \sin^2 \theta = \frac{16}{16} - \frac{15}{16} \] \[ \sin^2 \theta = \frac{16 - 15}{16} \] \[ \sin^2 \theta = \frac{1}{16} \] \[ \implies \sin \theta = \pm \frac{1}{4} \] Since x is in the IInd quadrant where sine is positive: \[ \therefore \sin x = \frac{1}{4} \]
In simple words: Use the Pythagorean identity and substitute the given cosine value. The angle range tells you that x is in the second quadrant, so sine must be positive. Take the positive square root to get your final answer.

Exam Tip: Always use the given angle range to determine which quadrant you are in - this tells you the sign of your answer before you even compute it. Never ignore the quadrant information.

Question 7. If sec x = -2 and \( \pi < x < \frac{3\pi}{2} \), find the values of all the other five trigonometric functions.
Answer: Given that sec x = -2. Since x lies in the third quadrant (because \( \pi < x < \frac{3\pi}{2} \)), both sine and cosine are negative, while tangent remains positive.
First, we calculate cos x:
\( \cos x = \frac{1}{\sec x} = \frac{1}{-2} = -\frac{1}{2} \)
Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\( \left(-\frac{1}{2}\right)^2 + \sin^2 x = 1 \)
\( \frac{1}{4} + \sin^2 x = 1 \)
\( \sin^2 x = \frac{3}{4} \)
\( \sin x = \pm\frac{\sqrt{3}}{2} \)
Since x is in the third quadrant where sine is negative:
\( \sin x = -\frac{\sqrt{3}}{2} \)
Now we find the remaining functions:
\( \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \)
\( \csc x = \frac{1}{\sin x} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} \)
\( \cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} \)
Therefore, the six trigonometric function values are shown in the table:

In simple words: When you know one trigonometric function value and the angle's quadrant, you can find all the others using basic identities and the signs that apply in that quadrant.

Exam Tip: Always identify the quadrant first to get the correct signs. The identity \( \sin^2 x + \cos^2 x = 1 \) is your main tool for finding missing values from just one function.

Question 8. Find the value of \( \sin\left(\frac{31\pi}{3}\right) \)
Answer: To find the value of \( \sin\frac{31\pi}{3} \):
\( \sin\frac{31\pi}{3} = \sin\left(10\pi + \frac{\pi}{3}\right) \)
\( = \sin\left(5 \times (2\pi) + \frac{\pi}{3}\right) \)
Since the sine function repeats after an interval of \( 2\pi \), we can ignore \( 5 \times (2\pi) \):
\( = \sin\left(\frac{\pi}{3}\right) \)
\( = \sin\left(\frac{1}{3} \times 180°\right) \)
\( = \sin 60° \)
\( = \frac{\sqrt{3}}{2} \)
In simple words: Since sine has a period of \( 2\pi \), remove all complete cycles from the angle. What's left is an angle from the first period whose sine value we can easily calculate.

Exam Tip: Use periodicity to reduce large angles. Always factor the angle as a multiple of the period plus a remainder that falls within \( 0 \) to \( 2\pi \).

Question 9. Find the value of \( \cos\left(\frac{17\pi}{2}\right) \)
Answer: To find the value of \( \cos\frac{17\pi}{2} \):
\( \cos\frac{17\pi}{2} = \cos\left(8\pi + \frac{\pi}{2}\right) \)
\( = \cos\left(4 \times (2\pi) + \frac{\pi}{2}\right) \)
Since cosine repeats after an interval of \( 2\pi \), we ignore \( 4 \times (2\pi) \):
\( = \cos\left(\frac{\pi}{2}\right) \)
\( = \cos\left(\frac{1}{2} \times 180°\right) \)
\( = \cos 90° \)
\( = 0 \)
In simple words: Pull out all the \( 2\pi \) cycles that you can, then evaluate the remaining angle.

Exam Tip: Cosine of \( 90° \) (or \( \frac{\pi}{2} \)) is always zero - this is a key value to memorize.

Question 10. Find the value of \( \tan\left(\frac{-25\pi}{3}\right) \)
Answer: To find the value of \( \tan\frac{-25\pi}{3} \):
We use the property \( \tan(-\theta) = -\tan\theta \):
\( \tan\left(\frac{-25\pi}{3}\right) = -\tan\left(\frac{25\pi}{3}\right) \)
\( = -\tan\left(8\pi + \frac{\pi}{3}\right) \)
\( = -\tan\left(4 \times (2\pi) + \frac{\pi}{3}\right) \)
Since tangent has a period of \( 2\pi \), we remove \( 4 \times (2\pi) \):
\( = -\tan\left(\frac{\pi}{3}\right) \)
\( = -\tan\left(\frac{1}{3} \times 180°\right) \)
\( = -\tan 60° \)
\( = -\sqrt{3} \)
In simple words: For negative angles, apply the odd function property first. Then reduce using the period.

Exam Tip: Remember that tangent is an odd function - \( \tan(-\theta) \) always equals \( -\tan\theta \). Apply this before removing cycles.

Question 11. Find the value of \( \cot\left(\frac{13\pi}{4}\right) \)
Answer: To find the value of \( \cot\frac{13\pi}{4} \):
We have:
\( \cot\frac{13\pi}{4} \)
Putting \( \pi = 180° \):
\( = \cot\left(\frac{13 \times 180°}{4}\right) \)
\( = \cot(13 \times 45°) \)
\( = \cot(585°) \)
\( = \cot(90° \times 6 + 45°) \)
\( = \cot 45° \)
Since \( 585° \) is in the third quadrant and the multiple of \( 90° \) is even:
\( = 1 \)
In simple words: Convert radians to degrees using \( \pi = 180° \), then reduce the angle by removing complete \( 360° \) cycles (or \( 90° \) cycles for periodicity analysis).

Exam Tip: When working with cotangent, note that \( \cot 45° = 1 \) is a standard value. Track which quadrant your reduced angle falls in to ensure the correct sign.

Question 12. Find the value of \( \sec\left(\frac{-25\pi}{3}\right) \)
Answer: To find the value of \( \sec\left(-\frac{25\pi}{3}\right) \):
We use the property \( \sec(-\theta) = \sec\theta \):
\( \sec\left(-\frac{25\pi}{3}\right) = \sec\frac{25\pi}{3} \)
Putting \( \pi = 180° \):
\( = \sec\frac{25 \times 180}{3} \)
\( = \sec[25 \times 60°] \)
\( = \sec[1500°] \)
\( = \sec[90° \times 16 + 60°] \)
Since \( 1500° \) is in the first quadrant and the multiple of \( 90° \) is even:
\( = \sec 60° \)
\( = 2 \)
In simple words: Secant is an even function - \( \sec(-\theta) \) equals \( \sec\theta \). Then reduce the angle and find the secant value.

Exam Tip: Secant of \( 60° \) equals 2 - this is a core reciprocal identity value. Always apply the even/odd property before angle reduction.

Question 13. Find the value of \( \csc\left(\frac{-41\pi}{4}\right) \)
Answer: To find the value of \( \csc\left(-\frac{41\pi}{4}\right) \):
We use the property \( \csc(-\theta) = -\csc\theta \):
\( \csc\left(-\frac{41\pi}{4}\right) = -\csc\frac{41\pi}{4} \)
Putting \( \pi = 180° \):
\( = -\csc\frac{41 \times 180}{4} \)
\( = -\csc[41 \times 45°] \)
\( = -\csc[1845°] \)
\( = -\csc[90° \times 20 + 45°] \)
Since \( 1845° \) is in the first quadrant and the multiple of \( 90° \) is even:
\( = -\csc 45° \)
\( = -\sqrt{2} \)
In simple words: Cosecant is an odd function, so put the negative sign out front first. Then reduce the angle and evaluate.

Exam Tip: Cosecant of \( 45° \) equals \( \sqrt{2} \) - memorize this value as it appears frequently. Don't forget to carry through the negative sign from the odd function property.

Question 14. Find the value of \( \sin 405° \)
Answer: To find the value of \( \sin 405° \):
\( \sin 405° = \sin[90° \times 4 + 45°] \)
\( = \sin 45° \)
Since \( 405° \) is in the first quadrant and the multiple of \( 90° \) is even:
\( = \frac{1}{\sqrt{2}} \)
In simple words: Remove complete \( 360° \) cycles (or analyze using \( 90° \) multiples) to find an equivalent angle in the first period, then calculate its sine.

Exam Tip: Sine of \( 45° \) is \( \frac{1}{\sqrt{2}} \) or \( \frac{\sqrt{2}}{2} \) - both forms are equivalent. Always reduce angles to the range \( [0°, 360°) \) first.

Question 15. Find the value of \( \sec(-1470°) \)
Answer: To find the value of \( \sec(-1470°) \):
We use the property \( \sec(-\theta) = \sec\theta \):
\( \sec(-1470°) = \sec(1470°) \)
\( = \sec[90° \times 16 + 30°] \)
Since \( 1470° \) is in the first quadrant and the multiple of \( 90° \) is even:
\( = \sec 30° \)
\( = \frac{2}{\sqrt{3}} \)
In simple words: Apply the even function property to remove the negative sign. Then use periodicity to find the coterminal angle in the standard range.

Exam Tip: Secant of \( 30° \) is \( \frac{2}{\sqrt{3}} \) - this is the reciprocal of cosine of \( 30° \). Always apply symmetry properties before reducing the angle.

Question 16. Find the value of \( \tan(-300°) \)
Answer: To find the value of \( \tan(-300°) \):
We use the property \( \tan(-\theta) = -\tan\theta \):
\( \tan(-300°) = -\tan(300°) \)
\( = -\tan[90° \times 3 + 30°] \)
Since \( 300° \) is in the fourth quadrant and the multiple of \( 90° \) is odd:
\( = -\cot 30° \)
\( = -\sqrt{3} \)
In simple words: Tangent is an odd function, so handle the negative sign first. Then reduce to a standard angle and evaluate.

Exam Tip: When the multiple of \( 90° \) is odd, the trigonometric function changes (sine to cosine, tangent to cotangent, etc.). Track this carefully to avoid sign errors.

Question 17. Find the value of \( \cot(585°) \)
Answer: To find the value of \( \cot(585°) \):
\( \cot(585°) = \cot[90° \times 6 + 45°] \)
\( = \cot 45° \)
Since \( 585° \) is in the third quadrant and the multiple of \( 90° \) is even:
\( = 1 \)
In simple words: Factor the angle as a multiple of \( 90° \) plus a remainder. When the multiple is even, the function stays the same.

Exam Tip: Cotangent of \( 45° \) is always 1. Use the \( 90° \) factorization method to determine both the reduced angle and which trigonometric function appears in the final answer.

Question 18. Find the value of \( \csc(-750°) \)
Answer: To find the value of \( \csc(-750°) \):
We use the property \( \csc(-\theta) = -\csc\theta \):
\( \csc(-750°) = -\csc(750°) \)
\( = -\csc[90° \times 8 + 30°] \)
Since \( 750° \) is in the first quadrant (as 405 is in the Ist quadrant) and the multiple of \( 90° \) is even:
\( = -\csc 30° \)
\( = -2 \)
In simple words: Cosecant is an odd function. Apply the odd property, then factor using \( 90° \) to find the coterminal angle.

Exam Tip: Cosecant of \( 30° \) is 2. The negative sign from the odd function property must be carried through to the final answer.

Question 19. Find the value of \( \cos(-2220°) \)
Answer: To find the value of \( \cos(-2220°) \):
We use the property \( \cos(-\theta) = \cos\theta \):
\( \cos(-2220°) = \cos 2220° \)
\( = \cos[2160° + 60°] \)
\( = \cos[360° \times 6 + 60°] \)
Since \( 2220° \) is in the first quadrant and the multiple of \( 360° \) is even:
\( = \cos 60° \)
\( = \frac{1}{2} \)
In simple words: Cosine is an even function, so the negative sign has no effect. Remove complete rotations (multiples of \( 360° \)) to find the standard position angle.

Exam Tip: Cosine of \( 60° \) is \( \frac{1}{2} \). For very large angles, subtract multiples of \( 360° \) until you get an angle between \( 0° \) and \( 360° \).

Question 20. Prove that \( \tan^2\frac{\pi}{3} + 2\cos^2\frac{\pi}{4} + 3\sec^2\frac{\pi}{6} + 4\cos^2\frac{\pi}{2} = 8 \)
Answer: To prove: \( \tan^2\frac{\pi}{3} + 2\cos^2\frac{\pi}{4} + 3\sec^2\frac{\pi}{6} + 4\cos^2\frac{\pi}{2} = 8 \)
Starting with the left side:
\( = \tan^2\frac{\pi}{3} + 2\cos^2\frac{\pi}{4} + 3\sec^2\frac{\pi}{6} + 4\cos^2\frac{\pi}{2} \)
Putting \( \pi = 180° \):
\( = \tan^2 60° + 2\cos^2 45° + 3\sec^2 30° + 4\cos^2 90° \)
Now, we know:
\( \tan 60° = \sqrt{3} \)
\( \cos 45° = \frac{1}{\sqrt{2}} \)
\( \sec 30° = \frac{2}{\sqrt{3}} \)
\( \cos 90° = 0 \)
Putting these values:
\( = (\sqrt{3})^2 + 2 \times \left(\frac{1}{\sqrt{2}}\right)^2 + 3 \times \left(\frac{2}{\sqrt{3}}\right)^2 + 4(0)^2 \)
\( = 3 + 2 \times \frac{1}{2} + 3 \times \frac{4}{3} + 0 \)
\( = 3 + 1 + 4 \)
\( = 8 \)
\( = \) RHS
\( \therefore \) LHS = RHS
Hence Proved
In simple words: Convert all angles to degrees or use known values. Substitute into each term, compute squares and products, then add them all together.

Exam Tip: Memorize the standard angle values (\( 30°, 45°, 60° \)). Always substitute values methodically, one term at a time, to avoid calculation errors.

Question 21. Prove that \( \sin\frac{\pi}{6}\cos 0 + \sin\frac{\pi}{4}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\cos\frac{\pi}{6} = \frac{7}{4} \)
Answer: To prove: \( \sin\frac{\pi}{6}\cos 0 + \sin\frac{\pi}{4}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\cos\frac{\pi}{6} = \frac{7}{4} \)
Starting with the left side:
\( = \sin\frac{\pi}{6}\cos 0 + \sin\frac{\pi}{4}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\cos\frac{\pi}{6} \)
Putting \( \pi = 180° \):
\( = \sin 30° \cos 0° + \sin 45° \cos 45° + \sin 60° \cos 30° \)
Now, we know:
\( \sin 30° = \frac{1}{2} \)
\( \cos 0° = 1 \)
\( \sin 45° = \frac{1}{\sqrt{2}} \)
\( \cos 45° = \frac{1}{\sqrt{2}} \)
\( \sin 60° = \frac{\sqrt{3}}{2} \)
\( \cos 30° = \frac{\sqrt{3}}{2} \)
Putting the values:
\( = \frac{1}{2} \times 1 + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \)
\( = \frac{1}{2} + \frac{1}{2} + \frac{3}{4} \)
\( = \frac{2 + 2 + 3}{4} \)
\( = \frac{7}{4} \)
\( = \) RHS
\( \therefore \) LHS = RHS
Hence Proved
In simple words: Look up each function value at the given angle, multiply pairs together as shown, add all three products, and verify the result.

Exam Tip: When multiplying two identical function values (like \( \sin 45° \times \cos 45° \)), you get a perfect square. Use this to simplify arithmetic quickly.

Question 22. Prove that \( 4\sin\frac{\pi}{6}\sin^2\frac{\pi}{3} + 3\cos\frac{\pi}{3}\tan\frac{\pi}{4} + \cos\text{ec}^2\frac{\pi}{2} = 4 \)
Answer: To prove: \( 4\sin\frac{\pi}{6}\sin^2\frac{\pi}{3} + 3\cos\frac{\pi}{3}\tan\frac{\pi}{4} + \csc^2\frac{\pi}{2} = 4 \)
Starting with the left side:
\( = 4\sin\frac{\pi}{6}\sin^2\frac{\pi}{3} + 3\cos\frac{\pi}{3}\tan\frac{\pi}{4} + \csc^2\frac{\pi}{2} \)
Putting \( \pi = 180° \):
\( = 4\sin 30° \sin^2 60° + 3\cos 60° \tan 45° + \csc^2 90° \)
Now, we know:
\( \sin 30° = \frac{1}{2} \)
\( \sin 60° = \frac{\sqrt{3}}{2} \)
\( \cos 60° = \frac{1}{2} \)
\( \tan 45° = 1 \)
\( \csc 90° = 1 \)
Putting the values:
\( = 4 \times \frac{1}{2} \times \left(\frac{\sqrt{3}}{2}\right)^2 + 3 \times \frac{1}{2} \times 1 + (1)^2 \)
\( = 2 \times \frac{3}{4} + \frac{3}{2} + 1 \)
\( = \frac{3}{2} + \frac{3}{2} + 1 \)
\( = \frac{3 + 3 + 2}{2} \)
\( = 4 \)
\( = \) RHS
\( \therefore \) LHS = RHS
Hence Proved
In simple words: Substitute standard angle values into each term. Square any functions that need squaring. Add all three results together.

Exam Tip: Watch for squared terms - they change the value significantly. Always compute squares before multiplying and adding to the rest of the expression.

Exercise 15(B)

Question 1. Find the value of
(i) cos 840°
(ii) sin 870°
(iii) tan (- 120°)
(iv) sec (- 420°)
(v) cosec (- 690°)
(vi) tan (225°)
(vii) cot (- 315°)
(viii) sin (- 1230°)
(ix) cos (495°)
Answer:
(i) \( \cos 840° = \cos(2.360° + 120°) \)

(using \( \cos(2a + x) = \cos x \))

\( = \cos(120°) \)

\( = \cos(180° - 60°) \)

\( = - \cos 60° \)

(using \( \cos(\omega - x) = - \cos x \))

\( = - \frac{1}{2} \)

(ii) \( \sin 870° = \sin(2.360° + 150°) \)

(using \( \sin(2a + x) = \sin x \))

\( = \sin 150° \)

\( = \sin(180° - 30°) \)

(using \( \sin(\omega - x) = \sin x \))

\( = \sin 30° \)

\( = \frac{1}{2} \)

(iii) \( \tan(- 120°) = - \tan 12 \)

(using \( \tan(- x) = \tan x \))

\( = - \tan(180° - 60°) \)

(in II quadrant tan is negative)

\( = - (- \tan 60°) \)

\( = \tan 60° \)

\( = \sqrt{3} \)

(iv) \( \sec(- 420°) = \frac{1}{\cos(- 420°)} \)

\( = \frac{1}{- \cos 420°} \)

(using \( \cos(- x) = - \cos x \))

\( = \frac{-1}{- \cos(360° + 60)} \)

(using \( \cos(2a + x) = \cos x \))

\( = \frac{-1}{\cos 60°} \Rightarrow \frac{-1}{1/2} = -2 \)

(v) \( \cosec(690°) = \frac{1}{\sin(- 690°)} \Rightarrow \frac{1}{-\sin(690°)} = \frac{1}{-\sin(2.360 - 30°)} \)

(IV quadrant sinx is negative)

\( = \frac{1}{-(-\sin 30°)} \Rightarrow \frac{1}{1/2} = 2 \)

(vi) \( \tan 225° = \tan(180° + 45°) \)

(in III quadrant tanx is positive)

\( \Rightarrow \tan 45° = 1 \)

(vii) \( \cot(- 315°) = \frac{1}{\tan(- 315)°} \Rightarrow \frac{1}{-\tan(315°)} = \frac{1}{-\tan(360° - 45°)} \)

(using \( \tan(- x) = - \tan x \))

\( = \frac{1}{-(-\tan 45°)} \Rightarrow 1 \)

(in IV quadrant tanx is negative)

(viii) \( \sin(- 1230°) = \sin 1230° \)

(using \( \sin(- x) = \sin x \))

\( = \sin(3.360° + 150°) \)

\( = \sin 150° \)

\( = \sin(180° - 30°) \)

(using \( \sin(180° - x) = \sin x \))

\( = \sin 30° \)

\( = \frac{1}{2} \)

(ix) \( \cos 495° = \cos(360° + 135°) \)

(using \( \cos(360° + x) = \cos x \))

\( = \cos 135° \)

\( = \cos(180° - 45°) \)

(using \( \cos(180° - x) = - \cos x \))

\( = - \cos 45° \)

\( = - \frac{1}{\sqrt{2}} \)

In simple words: To find trigonometric values of large angles, reduce them to standard angles (like 30°, 45°, 60°) by subtracting multiples of 360° or by using angle-difference identities. Then apply the quadrant rules to determine if the result is positive or negative.

Exam Tip: Always identify which quadrant your angle falls into after reduction - this tells you the sign of the trigonometric value. Memorize the angle reduction formulas and the ASTC rule for quick solving.

Question 2. Find the values of all trigonometric functions of 135°
Answer:
\( \sin 135° = \sin(180° - 45°) \)

(using \( \sin(180° - x) = \sin x \))

\( = \sin 45° \Rightarrow \frac{1}{\sqrt{2}} \)

\( \cos 135° = \cos(180° - 45°) \)

(using \( \cos(180° - x) = - \cos x \))

\( = \cos 45° \Rightarrow - \frac{1}{\sqrt{2}} \)

\( \tan 135° = \frac{\sin 135°}{\cos 135°} \Rightarrow \frac{1/\sqrt{2}}{-1/\sqrt{2}} = -1 \)

\( \csc 135° = \frac{1}{\sin 135°} \Rightarrow \sqrt{2} \)

\( \sec 135° = \frac{1}{\cos 135°} \Rightarrow - \sqrt{2} \)

\( \cot 135° = \frac{1}{\tan 135°} \Rightarrow -1 \)

In simple words: Break 135° into \( 180° - 45° \), then apply angle-difference rules. Sine stays positive in the second quadrant, but cosine becomes negative. Use these to calculate the remaining four functions.

Exam Tip: For angles in the second quadrant, sine is positive and all other functions are negative. Use the reference angle (45° here) to find the absolute values, then assign signs based on the quadrant.

Question 3. Prove that
(i) \( \sin 80° \cos 20° - \cos 80° \sin 20° = \frac{\sqrt{3}}{2} \)
(ii) \( \cos 45° \cos 15° - \sin 45° \sin 15° = \frac{1}{2} \)
(iii) \( \cos 75° \cos 15° + \sin 75° \sin 15° = \frac{1}{2} \)
(iv) \( \sin 40° \cos 20° + \cos 40° \sin 20° = \frac{\sqrt{3}}{2} \)
(v) \( \cos 130° \cos 40° + \sin 130° \sin 40° = 0 \)
Answer:
(i) \( \sin 80° \cos 20° - \cos 80° \sin 20° = \sin(80° - 20°) \)

(using \( \sin(A - B) = \sin A \cos B - \cos A \sin B \))

\( = \sin 60° \)

\( = \frac{\sqrt{3}}{2} \)

(ii) \( \cos 45° \cos 15° - \sin 45° \sin 15° = \cos(45° + 15°) \)

(using \( \cos(A + B) = \cos A \cos B - \sin A \sin B \))

\( = \cos 60° \)

\( = \frac{1}{2} \)

(iii) \( \cos 75° \cos 15° + \sin 75° \sin 15° = \cos(75° - 15°) \)

(using \( \cos(A - B) = \cos A \cos B + \sin A \sin B \))

\( = \cos 60° \)

\( = \frac{1}{2} \)

(iv) \( \sin 40° \cos 20° + \cos 40° \sin 20° = \sin(40° + 20°) \)

(using \( \sin(A + B) = \sin A \cos B + \cos A \sin B \))

\( = \sin 60° \)

\( = \frac{\sqrt{3}}{2} \)

(v) \( \cos 130° \cos 40° + \sin 130° \sin 40° = \cos(130° - 40°) \)

(using \( \cos(A - B) = \cos A \cos B + \sin A \sin B \))

\( = \cos 90° \)

\( = 0 \)

In simple words: Recognize which sum or difference formula matches each expression. Apply the formula to combine the angles into one, then evaluate the resulting standard angle.

Exam Tip: Always look for the pattern of products in addition/subtraction form - this helps you spot which formula to use. Memorize all four addition formulas (sin(A±B) and cos(A±B)) for quick recognition.

Question 4. Prove that
(i) \( \sin(50° + \theta) \cos(20° + \theta) - \cos(50° + \theta) \sin(20° + \theta) = \frac{1}{2} \)
(ii) \( \cos(70° + \theta) \cos(10° + \theta) + \sin(70° + \theta) \sin(10° + \theta) = \frac{1}{2} \)
Answer:
(i) \( \sin(50° + \theta) \cos(20° + \theta) - \cos(50° + \theta) \sin(20° + \theta) \)

\( = \sin(50° + \theta - (20° + \theta)) \)

(using \( \sin(A - B) = \sin A \cos B - \cos A \sin B \))

\( = \sin(50° + \theta - 20° - \theta) \)

\( = \sin 30° \)

\( = \frac{1}{2} \)

(ii) \( \cos(70° + \theta) \cos(10° + \theta) + \sin(70° + \theta) \sin(10° + \theta) \)

\( = \cos(70° + \theta - (10° + \theta)) \)

(using \( \cos(A - B) = \cos A \cos B + \sin A \sin B \))

\( = \cos(70° + \theta - 10° - \theta) \)

\( = \cos 60° \)

\( = \frac{1}{2} \)

In simple words: Even though the expressions have a variable \( \theta \), the addition formulas still work. When you subtract the two angles, the \( \theta \) cancels out, leaving only the difference of the constant parts.

Exam Tip: When angles contain variables or extra terms, apply the formula mechanically - the variables will often cancel, simplifying the problem significantly.

Question 5. Prove that
(i) \( \cos(n + 2)x \cos(n + 1)x + \sin(n + 2)x \sin(n + 1)x = \cos x \)
(ii) \( \cos\left(\frac{\pi}{4} - x\right) \cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right) \sin\left(\frac{\pi}{4} - y\right) = \sin(x + y) \)
Answer:
(i) \( \cos(n + 2)x \cos(n + 1)x + \sin(n + 2)x \sin(n + 1)x \)

\( = \sin((n + 2)x + (n + 1)x) \)

(using \( \cos(A - B) = \cos A \cos B + \sin A \sin B \))

\( = \cos(nx + 2x - (nx + x)) \)

\( = \cos(nx + 2x - nx - x) \)

\( = \cos x \)

(ii) \( \cos\left(\frac{\pi}{4} - x\right) \cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right) \sin\left(\frac{\pi}{4} - y\right) \)

\( = \cos\left(\frac{\pi}{4} - x + \frac{\pi}{4} - y\right) \)

(using \( \cos(A + B) = \cos A \cos B - \sin A \sin B \))

\( = \cos\left(\frac{2\pi}{4} - x - y\right) \)

\( = \cos\left(\frac{\pi}{2} - (x + y)\right) \)

(using \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \))

\( = \sin(x + y) \)

In simple words: Apply the sum and difference formulas to combine multiple angles with variables. Use the co-function identity \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta \) to convert cosine to sine when needed.

Exam Tip: Look for patterns in the angles - if they share a common structure, the formula will simplify them. Always check for complementary angles (those that add to \( \frac{\pi}{2} \)) to apply co-function identities.

Question 6. Prove that \( \frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} - x\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2 \)
Answer:
\( \frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} - x\right)} = \frac{\frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \cdot \tan x}}{\frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \cdot \tan x}} \)

\( \Rightarrow \frac{\frac{1 + \tan x}{1 - 1.\tan x}}{\frac{1 - \tan x}{1 + 1.\tan x}} = \frac{1 + \tan x}{1 - \tan x} \cdot \frac{1 + \tan x}{1 - \tan x} \)

\( \Rightarrow \left(\frac{1 + \tan x}{1 - \tan x}\right)^2 \)

Hence, Proved.

In simple words: Apply the tangent addition formula to both the numerator and denominator separately. Since \( \tan\frac{\pi}{4} = 1 \), the formulas simplify nicely. When you divide the two results, the fractions invert and multiply, giving you the square of one fraction.

Exam Tip: When you see compound angles involving \( \frac{\pi}{4} \), remember that \( \tan\frac{\pi}{4} = 1 \), which makes the addition formulas much simpler. Division of fractions means multiply by the reciprocal - use this to rearrange and simplify.

Question 7. Prove that
(i) \( \sin 75° = \frac{(\sqrt{6} + \sqrt{2})}{4} \)
(ii) \( \frac{\cos 135° - \cos 120°}{\cos 135° + \cos 120°} = (3 - 2\sqrt{2}) \)
(iii) \( \tan 15° + \cot 15° = 4 \)
Answer:
(i) \( \sin 75° = \sin(90° - 15°) \)

(using \( \sin(A - B) = \sin A \cos B - \cos A \sin B \))

\( = \sin 90° \cos 15° - \cos 90° \sin 15° \)

\( = 1. \cos 15° - 0. \sin 15° \)

\( = \cos 15° \)

\( \cos 15° = \cos(45° - 30°) \)

(using \( \cos(A - B) = \cos A \cos B + \sin A \sin B \))

\( = \cos 45°. \cos 30° + \sin 45°. \sin 30° \)

\( = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \)

\( = \frac{\sqrt{3} + 1}{2\sqrt{2}} .1 \Rightarrow \frac{\sqrt{3} + 1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} \)

\( \sin 75° = \cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4} \)

(ii) \( \frac{\cos 135° - \cos 120°}{\cos 135° + \cos 120°} = \frac{\cos(180° - 45°) - \cos(180° - 60°)}{\cos(180° - 45°) + \cos(180° - 60°)} \)

(using \( \sin(180° - x) \) = \( \sin x \))

\( = \sin x \))

(using \( \cos(180° - x) = - \cos x \))

\( = \frac{-\cos 45° - (-\cos 60°)}{-\cos 45° + (-\cos 60°)} \)

\( = \frac{\cos 60° - \cos 45°}{-(\cos 60° + \cos 45°)} \)

\( = \frac{\frac{1}{2} - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \frac{1}{2}} \Rightarrow \frac{\frac{1 - \sqrt{2}}{2}}{\frac{\sqrt{2} + 1}{2}} = \frac{1 - \sqrt{2}}{\sqrt{2} + 1} = -\frac{1 - \sqrt{2}(-\sqrt{2} + 1)}{sqrt{2} + 1(-\sqrt{2} + 1)} \)

\( = - \frac{-\sqrt{2} + 1 + 2 - \sqrt{2}}{-2 + \sqrt{2} - \sqrt{2} + 1} \Rightarrow - \frac{-2\sqrt{2} + 3}{-1} = 3 - 2\sqrt{2} \)

(iii) \( \tan 15° + \cot 15° = \)

First, calculate \( \tan 15° \):

\( \tan 15° = \frac{\sin 15°}{\cos 15°} \)

\( [\cos 15° = \frac{\sqrt{3} + 1}{2\sqrt{2}}, \sin 15° = \sin(45° - 30°) \)

\( = \sin 45°. \cos 30° - \cos 45°. \sin 30° = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \)

\( = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

\( \tan 15° = \frac{\frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}} \Rightarrow \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)

and \( \cot 15° = \frac{1}{\tan 15°} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)

Putting in eq(1),

\( \tan 15° + \cot 15° = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)

\( = \frac{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3}}{2} \)

\( = \frac{8}{2} = 4 \)

In simple words: For part (i), break 75° into simpler angles using the difference formula. For part (ii), use the supplementary angle relationships to express larger angles in terms of standard ones, then substitute and simplify. For part (iii), find tan 15° and cot 15° separately by breaking 15° into 45° - 30°, then add them using a common denominator.

Exam Tip: Always look for ways to express non-standard angles as sums or differences of standard angles (30°, 45°, 60°). Rationalize denominators and combine fractions carefully - small algebra errors can lead to wrong final answers.

Question 8. Prove that
(i) \( \cos 15° - \sin 15° = \frac{1}{\sqrt{2}} \)
(ii) \( \cot 105° - \tan 105° = 2\sqrt{3} \)
(iii) \( \frac{\tan 69° + \tan 66°}{1 - \tan 69° \tan 66°} = -1 \)
Answer:
(i) \( \cos 15° = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

\( \sin 15° = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

\( \cos 15° - \sin 15° = \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

\( = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2\sqrt{2}} \)

\( = \frac{2}{2\sqrt{2}} \)

\( = \frac{1}{\sqrt{2}} \)

(ii) \( \cot 105° - \tan 105° = \cot(180° - 75°) - \tan(180° - 75°) \)

(II quadrant tanx is negative and cotx as well)

\( = - \cot 75° - (- \tan 75°) \)

\( = \tan 75° - \cot 75° \)

\( \tan 75° = \frac{\sin 75°}{\cos 75°} \Rightarrow \frac{\sin(90° - 15°)}{\cos(90° - 15°)} = \frac{-\cos 15°}{\sin 15°} \)

(using \( \sin(90° - x) = - \cos x \) and \( \cos(90° - x) = \sin x \))

\( = - \frac{\sqrt{3} + 1}{2\sqrt{2}} \Rightarrow \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)

\( \cot 75° = \frac{1}{\tan 75°} \Rightarrow \frac{\sqrt{3} - 1}{-\sqrt{3} - 1} \)

\( \cot 105° - \tan 105° \)

\( = \frac{\sqrt{3} - 1}{-\sqrt{3} - 1} - \frac{\sqrt{3} - 1}{-\sqrt{3} - 1} \Rightarrow \frac{(\sqrt{3} - 1) - (-\sqrt{3} - 1)}{(-\sqrt{3} - 1)(-\sqrt{3} - 1)} = \frac{3 + 1 - 2\sqrt{3} - (3 + 1 + 2\sqrt{3})}{(-3 + 1 - \sqrt{3} + \sqrt{3})} \)

\( = \frac{-4\sqrt{3}}{-2} \Rightarrow 2\sqrt{3} \)

(iii) \( \frac{\tan 69° + \tan 66°}{1 - \tan 69°. \tan 66°} = \tan(69° + 66°) \Rightarrow \tan 135° = \tan(180° - 45°) \)

(II quadrant tanx negative)

\( \Rightarrow - \tan 45° = - 1 \)

In simple words: For part (i), use the known values of cos 15° and sin 15° from the previous question and subtract them directly. For part (ii), convert angles in the second quadrant to their acute angle equivalents and apply sign rules. For part (iii), recognize the tangent addition formula in the given expression and simplify using standard angles.

Exam Tip: Master the supplementary angle formulas (180° - x and 90° - x) - they help reduce angles in other quadrants to standard first-quadrant values. Always identify which formula pattern matches the problem structure before substituting numbers.

Question 9. Prove that \( \frac{\cos 9° + \sin 9°}{\cos 9° - \sin 9°} = \tan 54° \)
Answer: Start by factoring \( \cos 9° \) from both numerator and denominator.

\( \frac{\cos 9° + \sin 9°}{\cos 9° - \sin 9°} = \frac{\cos 9°(1 + \tan 9°)}{\cos 9°(1 - \tan 9°)} \Rightarrow \frac{\tan 45° + \tan 9°}{1 - \tan 45°. \tan 9°} = \tan(45° + 9°) \Rightarrow \tan 54 \)

(using \( \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x. \tan y} \) and \( \tan 45° = 1 \))

In simple words: Factor out \( \cos 9° \) from both the top and bottom. This reveals the tangent addition formula pattern. Since the denominator has 1 in the numerator formula, recognize that 1 comes from \( \tan 45° \), then apply the formula directly.

Exam Tip: When you see a fraction with cos and sin in both numerator and denominator, try dividing top and bottom by the same trig function (usually cos) to reveal a tangent addition formula. This transforms the problem into recognizing a standard formula pattern.

Question 10. Prove that \( \frac{\cos 8° - \sin 8°}{\cos 8° + \sin 8°} = \tan 37° \)
Answer: Start by factoring \( \cos 8° \) from both numerator and denominator.

\( \frac{\cos 8° - \sin 8°}{\cos 8° + \sin 8°} = \frac{\cos 8°(1 - \tan 8°)}{\cos 8°(1 + \tan 8°)} \Rightarrow \frac{\tan 45° - \tan 8°}{1 + \tan 45°. \tan 8°} = \tan(45° - 8°) \Rightarrow \tan 37° \)

[using \( \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x. \tan y} \) and \( \tan 45° = 1 \)]

In simple words: Factor \( \cos 8° \) from the numerator and denominator. The resulting expression matches the tangent difference formula, where the numerator structure gives tan 45° minus tan 8°. Apply the formula to get tan(45° - 8°) = tan 37°.

Exam Tip: For problems with subtraction in the numerator and addition in the denominator, use the tangent difference formula. Conversely, addition over subtraction signals the tangent addition formula. Memorize which goes where to identify the right formula instantly.

Question 11. Prove that \( \frac{\cos(\pi + \theta) \cos(- \theta)}{\cos(\pi - \theta) \cos\left(\frac{\pi}{2} + \theta\right)} = - \cot \theta \)
Answer:
\( \frac{\cos(\pi + \theta) \cos(- \theta)}{\cos(\pi - \theta) \cos\left(\frac{\pi}{2} + \theta\right)} = \frac{-\cos \theta. \cos \theta}{-\cos \theta. - \sin \theta} \)

\( \Rightarrow \frac{\cos \theta}{-\sin \theta} = - \cot \theta \)

[Using \( \cos(\pi - \theta) = -\cos \theta \) and \( \cos\left(\frac{\pi}{2} - \theta\right) = -\sin \theta, \cos(- \theta) = -\cos \theta \))]

(In III quadrant cosx is negative, \( \cos(\pi + \theta) = - \cos \theta \))

In simple words: Apply the angle transformation formulas to each trig function in the expression. \( \cos(\pi + \theta) \) becomes \( -\cos \theta \), \( \cos(- \theta) \) stays \( \cos \theta \), \( \cos(\pi - \theta) \) becomes \( -\cos \theta \), and \( \cos\left(\frac{\pi}{2} + \theta\right) \) becomes \( -\sin \theta \). Substitute all these and simplify by canceling common factors.

Exam Tip: Memorize the transformations for angles involving \( \pi \), \( \frac{\pi}{2} \), and negatives - these appear constantly. Write out all substitutions before simplifying to avoid sign errors in the final result.

Question 12. Prove that \( \frac{\cos \theta}{\sin(90° + \theta)} + \frac{\sin(- \theta)}{\sin(180° + \theta)} - \frac{\tan(90° + \theta)}{\cot \theta} = 3 \)
Answer: Using \( \sin(90° + \theta) = \cos \theta \) and \( \sin(- \theta) = \sin \theta \), \( \tan(90° + \theta) = - \cot \theta \)

\( \sin(180° + \theta) = - \sin \theta \) (III quadrant sinx is negative)

\( \frac{\cos \theta}{\sin(90° + \theta)} + \frac{\sin(- \theta)}{\sin(180° + \theta)} - \frac{\tan(90° + \theta)}{\cot \theta} = \frac{\cos \theta}{\cos \theta} + \frac{-\sin \theta}{-\sin \theta} - \frac{-\cot \theta}{\cot \theta} \)

\( = 1 + (1) - (-1) \Rightarrow 1 + 1 + 1 = 3 \)

In simple words: Substitute each of the transformed trig functions into the original expression. Each fraction simplifies to either 1 or -1, making the final addition straightforward.

Exam Tip: Always transform angles in terms of \( 90° + \theta \) and \( 180° + \theta \) first - these have predictable co-function and sign relationships. Once transformed, look for cancellation opportunities to simplify fractions quickly.

Question 13. Prove that \( \frac{\sin(180° + \theta) \cos(90° + \theta) \tan(270° - \theta) \cot(360° - \theta)}{\sin(360° - \theta) \cos(360° + \theta) \csc(- \theta) \sin(270° + \theta)} = 1 \)
Answer: Using \( \cos(90° + \theta) = - \sin \theta \) (I quadrant cosx is positive

\( \csc(- \theta) = - \csc \theta \)

\( \tan(270° - \theta) = \tan(180° + 90° - \theta) = \tan(90° - \theta) = \cot \theta \)

(III quadrant tanx is positive)

Similarly \( \sin(270° + \theta) = - \cos \theta \) (IV quadrant sinx is negative

\( \cot(360° - \theta) = \cot \theta \) (IV quadrant cotx is negative)

\( = \frac{\sin(180° + \theta) \cos(90° + \theta) \tan(270° - \theta) \cot(360° - \theta)}{\sin(360° - \theta) \cos(360° + \theta) \csc(- \theta) \sin(270° + \theta)} \)

\( = \frac{-\sin \theta. - \sin \theta. \cot \theta. - \cot \theta}{-\sin \theta. \cos \theta. - \csc \theta. - \cos \theta} \)

\( = \frac{-\sin \theta. - \sin \theta. \cot \theta. - \cot \theta}{-\sin \theta. \cos \theta. - \csc \theta. - \cos \theta} \)

\( = \cot \theta. \tan \theta. \cot \theta. \tan \theta \Rightarrow 1 \)

In simple words: Transform each angle using the quadrant rules and supplementary/co-function identities. Substitute the transformed values, then simplify by canceling terms that appear in both numerator and denominator. The product of cot and tan is always 1, so repeated pairs will cancel nicely.

Exam Tip: For complex multi-angle expressions, write out the transformation of each angle separately before substituting - this reduces the chance of sign errors. Always look for reciprocal pairs (like tan and cot, or sin and csc) that multiply to 1.

Question 14. If θ and Φ lie in the first quadrant such that sin θ = 8/17 and cos Φ = 12/13, find the values of
(i) sin (θ - Φ)
(ii) cos (θ - Φ)
(iii) tan (θ - Φ)
Answer: Given sin θ = 8/17 and cos Φ = 12/13

First, we determine cos θ and sin Φ.

\[ \cos \theta = \sqrt{1 - \sin^2 \theta} \Rightarrow \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{\frac{289 - 64}{289}} \Rightarrow \sqrt{\frac{225}{289}} = \frac{15}{17} \]

\[ \sin \Phi = \sqrt{1 - \cos^2 \Phi} \Rightarrow \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{169 - 144}{169}} = \sqrt{\frac{25}{169}} \Rightarrow \frac{5}{13} \]

(i) sin(θ - Φ) = sin θ cos Φ + cos θ sin Φ

\[ = \frac{8}{17} \cdot \frac{12}{13} + \frac{15}{17} \cdot \frac{5}{13} \Rightarrow \frac{96 + 75}{221} = \frac{171}{221} \]

(ii) cos(θ - Φ) = cos θ cos Φ + sin θ sin Φ

\[ = \frac{15}{17} \cdot \frac{12}{13} + \frac{8}{17} \cdot \frac{5}{13} \Rightarrow \frac{180 + 40}{221} = \frac{220}{221} \]

(iii) We first find tan θ and tan Φ.

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \Rightarrow \frac{8/17}{15/17} = \frac{8}{15} \quad \text{and} \quad \tan \Phi = \frac{\sin \Phi}{\cos \Phi} \Rightarrow \frac{5/13}{12/13} = \frac{5}{12} \]

\[ \tan(\theta - \Phi) = \frac{\tan \theta - \tan \Phi}{1 + \tan \theta \tan \Phi} \Rightarrow \frac{\frac{8}{15} - \frac{5}{12}}{1 + \frac{8}{15} \cdot \frac{5}{12}} = \frac{\frac{96 - 75}{180}}{\frac{180 + 40}{180}} = \frac{21}{220} \]
In simple words: Use the Pythagorean identity to locate the missing sine and cosine values from the given data. Then use angle difference formulas to find sin(θ - Φ), cos(θ - Φ), and tan(θ - Φ).

Exam Tip: Always check that both angles are in the first quadrant so all trigonometric ratios remain positive. Apply the angle difference formulas methodically to avoid arithmetic mistakes.

Question 15. If x and y are acute such that sin x = 1/√5 and sin y = 1/√10, prove that (x + y) = π/4
Answer: Given sin x = 1/√5 and sin y = 1/√10.

Now we compute the values of cos x and cos y.

\[ \cos x = \sqrt{1 - \sin^2 x} \Rightarrow \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{\frac{5 - 1}{5}} \Rightarrow \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]

\[ \cos y = \sqrt{1 - \sin x^2} \Rightarrow \sqrt{1 - \left(\frac{1}{\sqrt{10}}\right)^2} = \sqrt{\frac{10 - 1}{10}} \Rightarrow \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} \]

Sin(x + y) = sin x cos y + cos x sin y

\[ = \frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} \Rightarrow \frac{3 + 2}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \]

\[ \Rightarrow \sin(x + y) = \frac{1}{\sqrt{2}} \]

\[ \Rightarrow x + y = \frac{\pi}{4} \]
In simple words: Work out cos x and cos y from the given sin values. Then apply the sin addition formula. If the result equals 1/√2, the angle sum must be π/4.

Exam Tip: Recognise that \( \sin(\pi/4) = 1/\sqrt{2} \) is a standard value. This helps you quickly identify the angle from the final sine result.

Question 16. If x and y are acute angles such that cos x = 13/14 and cos y = 1/7, prove that (x - y) = -π/3
Answer: Given cos x = 13/14 and cos y = 1/7.

Now we compute the values of sin x and sin y.

\[ \sin x = \sqrt{1 - \cos^2 x} \Rightarrow \sqrt{1 - \left(\frac{13}{14}\right)^2} = \sqrt{\frac{196 - 169}{196}} \Rightarrow \sqrt{\frac{27}{196}} = \frac{3\sqrt{3}}{14} \]

\[ \sin y = \sqrt{1 - \cos^2 y} \Rightarrow \sqrt{1 - \left(\frac{1}{7}\right)^2} = \sqrt{\frac{49 - 1}{49}} \Rightarrow \sqrt{\frac{48}{49}} = \frac{4\sqrt{3}}{7} \]

Therefore,

Cos(x - y) = cos x cos y + sin x sin y

\[ = \frac{13}{14} \cdot \frac{1}{7} + \frac{3\sqrt{3}}{14} \cdot \frac{4\sqrt{3}}{7} \Rightarrow \frac{13 + 36}{98} = \frac{49}{98} = \frac{1}{2} \]

\[ \cos(x - y) = \frac{1}{2} \]

\[ x - y = \frac{\pi}{3} \]
In simple words: Find sin x and sin y using the Pythagorean relation. Apply the cosine difference formula. When the result equals 1/2, the angle difference is π/3.

Exam Tip: Be careful with signs and exact form of roots. Verify that both angles are in the allowable range before concluding the final angle relationship.

Question 17. If sin x = 12/13 and sin y = 4/5, where π/2 < x < π and 0 < y < π/2, find the values of
(i) sin (x + y)
(ii) cos (x + y)
(iii) tan (x - y)
Answer: Given sin x = 12/13 and sin y = 4/5

Here we compute the values of cos x and cos y.

\[ \cos x = \sqrt{1 - \sin^2 x} \Rightarrow \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{\frac{169 - 144}{169}} \Rightarrow \sqrt{\frac{25}{169}} = \frac{5}{13} \]

\[ \cos y = \sqrt{1 - \sin^2 y} \Rightarrow \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{\frac{25 - 16}{25}} \Rightarrow \sqrt{\frac{9}{25}} = \frac{3}{5} \]

(i) sin(x + y) = sin x cos y + cos x sin y

\[ \Rightarrow \frac{12}{13} \cdot \frac{3}{5} + \frac{5}{13} \cdot \frac{4}{5} \Rightarrow \frac{36 + 20}{65} = \frac{56}{65} \]

(ii) cos(x + y) = cos x cos y + sin x sin y

\[ = \frac{5}{13} \cdot \frac{3}{5} + \frac{12}{13} \cdot \frac{4}{5} \Rightarrow \frac{15 + 48}{65} = \frac{63}{65} \]

(iii) First we compute the values of tan x and tan y,

\[ \tan x = \frac{\sin x}{\cos x} \Rightarrow \frac{12/13}{5/13} = \frac{5}{12} \quad \text{and} \quad \tan y = \frac{\sin y}{\cos y} \Rightarrow \frac{4/5}{3/5} = \frac{4}{3} \]

\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \Rightarrow \frac{\frac{5}{12} - \frac{4}{3}}{1 + \frac{5}{12} \cdot \frac{4}{3}} = \frac{\frac{5 - 16}{12}}{\frac{36 + 20}{36}} = \frac{-11/12}{56/36} = \frac{-33}{56} \]
In simple words: Find cos x and cos y using the Pythagorean identity. Use addition and subtraction formulas for sine, cosine, and tangent to compute the required values.

Exam Tip: Notice the quadrant information - x is in the second quadrant where sine is positive but cosine is negative. Always use the quadrant to determine the correct sign of missing ratios.

Question 18. If cos x = 3/5 and cos y = -24/25, where 3π/2 < x < 2π and π < y < 3π/2, find the values of
(i) sin (x + y)
(ii) cos (x - y)
(iii) tan (x + y)
Answer: Given cos x = 3/5 and cos y = -24/25

We compute the values of sin x and sin y.

\[ \sin x = \sqrt{1 - \cos^2 x} \Rightarrow \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{25 - 9}{25}} \Rightarrow \sqrt{\frac{16}{25}} = \frac{4}{5} \]

\[ \sin y = \sqrt{1 - \cos^2 y} \Rightarrow \sqrt{1 - \left(\frac{-24}{25}\right)^2} = \sqrt{\frac{625 - 576}{625}} \Rightarrow \sqrt{\frac{49}{625}} = \frac{7}{25} \]

(i) sin(x + y) = sin x cos y + cos x sin y

\[ = \frac{4}{5} \cdot \frac{-24}{25} + \frac{3}{5} \cdot \frac{7}{25} \Rightarrow \frac{-96 + 21}{125} = \frac{-75}{125} = \frac{-3}{5} \]

(ii) cos(x - y) = cos x cos y + sin x sin y

\[ = \frac{3}{5} \cdot \frac{-24}{25} + \frac{4}{5} \cdot \frac{7}{25} \Rightarrow \frac{-72 + 28}{125} = \frac{-44}{125} \]

(iii) First we compute the values of tan x and tan y,

\[ \tan x = \frac{\sin x}{\cos x} \Rightarrow \frac{4/5}{3/5} = \frac{4}{3} \quad \text{and} \quad \tan y = \frac{\sin y}{\cos y} \Rightarrow \frac{7/25}{-24/25} = \frac{7}{-24} \]

\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \Rightarrow \frac{\frac{4}{3} + \frac{-7}{24}}{1 - \frac{4}{3} \cdot \frac{-7}{24}} = \frac{\frac{32 - 7}{24}}{\frac{72 + 28}{72}} = \frac{25/24}{100/72} = \frac{25}{24} \cdot \frac{72}{100} = \frac{75}{100} = \frac{3}{4} \]
In simple words: Determine sin x and sin y from the cosine values using the Pythagorean relation and the given quadrant information. Apply addition and subtraction formulas to find the required results.

Exam Tip: Quadrant constraints are crucial - x in the fourth quadrant has negative sine, and y in the third quadrant has negative sine. Always verify the correct sign before applying formulas.

Question 19. Prove that
(i) \( \cos\left(\frac{\pi}{3} + x\right) = \frac{1}{2}(\cos x - \sqrt{3} \sin x) \)

\[ \cos\left(\frac{\pi}{3} + x\right) = \cos\frac{\pi}{3}\cos x - \sin\frac{\pi}{3}\sin x \]

\[ \Rightarrow \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x = \frac{1}{2}(\cos x - \sqrt{3}\sin x) \]

(ii) \( \sin\left(\frac{\pi}{4} + x\right) + \sin\left(\frac{\pi}{4} - x\right) = \sqrt{2} \cos x \)

\[ = \sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x - \cos\frac{\pi}{4}\sin x \]

\[ = 2\sin\frac{\pi}{4}\cos x \Rightarrow 2 \cdot \frac{1}{\sqrt{2}}\cos x = \sqrt{2}\cos x \]

(iii) \( \frac{1}{\sqrt{2}}\cos\left(\frac{\pi}{4} + x\right) = \frac{1}{2}(\cos x - \sin x) \)

\[ \frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x\right) = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x\right) = \frac{1}{2}(\cos x - \sin x) \]

(iv) \( \cos x + \cos\left(\frac{2\pi}{3} + x\right) + \cos\left(\frac{2\pi}{3} - x\right) = 0 \)

\[ = \cos x + \cos\frac{2\pi}{3}\cos x - \sin\frac{2\pi}{3}\sin x + \cos\frac{2\pi}{3}\cos x + \sin\frac{2\pi}{3}\sin x \]

\[ = \cos x + 2\cos\left(\pi - \frac{\pi}{3}\right)\cos x = \cos x + 2\left(-\frac{1}{2}\right)\cos x = \cos x - \cos x \Rightarrow 0 \]
In simple words: Apply angle addition and subtraction formulas with standard angle values like π/3, π/4, and 2π/3. Simplify by combining like terms and using known trigonometric ratios.

Exam Tip: Memorise the sine and cosine values of standard angles (π/6, π/4, π/3, 2π/3). These recurring values speed up simplification significantly.

Question 20. Prove that
(i) \( 2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} = \frac{1}{2} \)

\[ 2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} = -\left(\cos\left(\frac{5\pi}{12} + \frac{\pi}{12}\right) - \cos\left(\frac{5\pi}{12} - \frac{\pi}{12}\right)\right) \]

\[ = -\left(\cos\frac{6\pi}{12} - \cos\frac{4\pi}{12}\right) = -\left(\cos\frac{\pi}{2} - \cos\frac{\pi}{3}\right) \Rightarrow -\left(0 - \frac{1}{2}\right) = \frac{1}{2} \]

(ii) \( 2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{1}{2} \)

\[ 2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} = \cos\left(\frac{5\pi}{12} + \frac{\pi}{12}\right) + \cos\left(\frac{5\pi}{12} - \frac{\pi}{12}\right) = \cos\frac{\pi}{2} + \cos\frac{\pi}{3} = 0 + \frac{1}{2} = \frac{1}{2} \]

(iii) \( 2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} = \frac{2 + \sqrt{3}}{2} \)

\[ 2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} = \sin\left(\frac{5\pi}{12} + \frac{\pi}{12}\right) + \sin\left(\frac{5\pi}{12} - \frac{\pi}{12}\right) \]

\[ = \sin\frac{6\pi}{12} + \sin\frac{4\pi}{12} \Rightarrow \sin\frac{\pi}{2} + \sin\frac{\pi}{3} = 1 + \frac{\sqrt{3}}{2} \Rightarrow \frac{2 + \sqrt{3}}{2} \]
In simple words: Use product-to-sum formulas (for converting products of sines and cosines into sums). Evaluate the resulting standard angles and simplify.

Exam Tip: Memorise the product-to-sum identities: they transform products into sums, which are often easier to evaluate using standard angle values.

Exercise 15C

Question 1. Prove that sin(150° + x) + sin(150° - x) = cos x
Answer: This proof uses the following formulas:

Sin(A + B) = sin A cos B + cos A sin B
Sin(A - B) = sin A cos B - cos A sin B

\[ = \sin150° \cos x + \cos150° \sin x + \sin150° \cos x - \cos150° \sin x \]

\[ = 2\sin150° \cos x = 2\sin(90° + 60°)\cos x = 2\cos60° \cos x = 2 \times \frac{1}{2} \cos x = \cos x \]
In simple words: Write out each sine expression using the angle sum formula. The terms with sin x cancel, leaving only those with cos x. Simplify using the fact that sin 150° = cos 60°.

Exam Tip: Recognise that sin(90° + A) = cos A - this transforms sin 150° into cos 60°, simplifying the calculation.

Question 2. Prove that cos x + cos(120° - x) + cos(120° + x) = 0
Answer: This proof uses the following formulas:

cos(A + B) = cos A cos B - sin A sin B
cos(A - B) = cos A cos B + sin A sin B

\[ = \cos x + \cos120° \cos x - \sin120\sin x + \cos120°\cos x + \sin120\sin x \]

\[ = \cos x + 2\cos120 \cos x = \cos x + 2\cos(90 + 30) \cos x = \cos x + 2(-\sin30) \cos x = \cos x - 2 \times \frac{1}{2} \cos x = \cos x - \cos x = 0 \]
In simple words: Apply the cosine addition formulas to both terms with 120°. The sine terms cancel, leaving only cosine terms. Use cos 120° = -1/2 to finish.

Exam Tip: Remember that cos(90° + A) = -sin A. This identity helps convert cos 120° into a simpler form.

Question 3. Prove that \( \sin\left(x - \frac{\pi}{6}\right) - \cos\left(x - \frac{\pi}{3}\right) = \sqrt{3} \sin x \)
Answer: This proof uses the following formulas:

sin(A - B) = sin A cos B - cos A sin B
cos(A - B) = cos A cos B + sin A sin B

\[ = \sin x \cos\frac{\pi}{6} - \cos x \sin\frac{\pi}{6} + \cos x \cos\frac{\pi}{3} + \sin x \sin\frac{\pi}{3} \]

\[ = \sin x \times \frac{\sqrt{3}}{2} - \cos x \times \frac{1}{2} + \cos x \times \frac{1}{2} + \sin x \times \frac{\sqrt{3}}{2} \]

\[ = \sin x \times \frac{\sqrt{3}}{2} + \sin x \times \frac{\sqrt{3}}{2} = \left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right) \sin x = \sqrt{3} \sin x \]
In simple words: Expand both angle expressions using the subtraction formulas. Substitute the standard angle values. The cosine terms cancel, leaving only the sine terms which sum to √3 sin x.

Exam Tip: Pay attention to the sign before the second term - it changes sin to a subtraction that becomes addition when you distribute the negative sign.

Question 4. Prove that \( \tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x} \)
Answer: This proof uses the following formula:

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \tan x} = \frac{1 + \tan x}{1 - \tan x} \cdot \tan\frac{\pi}{4} = 1 \]
In simple words: Use the tangent addition formula with A = π/4 and B = x. Since tan(π/4) = 1, substitute this value into the formula to get the required result.

Exam Tip: The tangent of π/4 is always 1 - this is a key value that simplifies many proofs involving this angle.

Question 5. Prove that \( \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} \)
Answer: This proof uses the following formula:

\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x} \cdot \tan\frac{\pi}{4} = 1 \]
In simple words: Apply the tangent subtraction formula with A = π/4 and B = x. Substitute tan(π/4) = 1 to reach the target form.

Exam Tip: This is the reciprocal-like counterpart to the previous question. Notice the pattern: addition and subtraction versions are related by swapping the positions and signs in the numerator and denominator.

Question 6. Express each of the following as a product.
(1) sin 10x + sin 6x
(2) sin 7x - sin 3x
(3) cos 7x + cos 5x
(4) cos 2x - cos 4x
Answer:

1. sin 10x + sin 6x = 2 sin \(\frac{10x + 6x}{2}\) cos \(\frac{10x - 6x}{2}\) = 2 sin \(\frac{18x}{2}\) cos \(\frac{4x}{2}\) = 2 sin 9x cos 2x

Using the formula:
sin(A + B) = sin A cos B + cos A sin B

2. sin 7x - sin 3x = 2 cos \(\frac{7x + 3x}{2}\) sin \(\frac{7x - 3x}{2}\) = 2 cos \(\frac{10x}{2}\) sin \(\frac{4x}{2}\) = 2 cos 5x sin 2x

Using the formula:
sin(A - B) = sin A cos B - cos A sin B

3. cos 7x + cos 5x = 2 cos \(\frac{7x + 5x}{2}\) cos \(\frac{7x - 5x}{2}\) = 2 cos \(\frac{12x}{2}\) cos \(\frac{2x}{2}\) = 2 cos 6x cos x

Using the formula:
cos(A + B) = cos A cos B - sin A sin B

4. cos 2x - cos 4x = -2 sin \(\frac{2x + 4x}{2}\) sin \(\frac{2x - 4x}{2}\) = -2 sin \(\frac{6x}{2}\) sin \(\frac{-2x}{2}\) = 2 sin 3x sin x

Using the formula:
cos(A - B) = cos A cos B + sin A sin B

In simple words: Use sum-to-product formulas to convert sums and differences of sines or cosines into products of trigonometric functions.

Exam Tip: Memorise the four sum-to-product identities - they are the reverse of product-to-sum formulas and appear frequently in simplification problems.

Question 7. Express each of the following as an algebraic sum of sines or cosines:
(i) 2 sin 6x cos 4x
(ii) 2 cos 5x sin 3x
(iii) 2 cos 7x cos 3x
(iv) 2 sin 8x sin 2x
Answer:

(i) 2 sin 6x cos 4x = sin(6x + 4x) + sin(6x - 4x) = sin 10x + sin 2x

Using the formula:
2 sin A cos B = sin(A + B) + sin(A - B)

(ii) 2 cos 5x sin 3x = sin(5x + 3x) - sin(5x - 3x) = sin 8x - sin 2x

Using the formula:
2 cos A sin B = sin(A + B) - sin(A - B)

(iii) 2 cos 7x cos 3x = cos(7x + 3x) + cos(7x - 3x) = cos 10x + cos 4x

Using the formula:
2 cos A cos B = cos(A + B) + cos(A - B)

(iv) 2 sin 8x sin 2x = cos(8x - 2x) - cos(8x + 2x) = cos 6x - cos 10x

Using the formula:
2 sin A sin B = cos(A - B) - cos(A + B)

In simple words: Apply product-to-sum formulas to rewrite each product as a sum or difference of trigonometric functions with different arguments.

Exam Tip: These product-to-sum identities are the inverse of the sum-to-product formulas. Knowing both directions helps with flexible problem solving.

Question 8. Prove that \( \frac{\sin x + \sin 3x}{\cos x - \cos 3x} = \cot x \)
Answer:

\[ \frac{\sin x + \sin 3x}{\cos x - \cos 3x} = \frac{2 \sin\frac{x + 3x}{2} \cos\frac{x - 3x}{2}}{-2 \sin\frac{x + 3x}{2} \sin\frac{x - 3x}{2}} \]

\[ = \frac{2 \sin\frac{4x}{2} \cos\frac{-2x}{2}}{-2 \sin\frac{4x}{2} \sin\frac{-2x}{2}} = \frac{\cos x}{\sin x} = \cot x \]

Using the formulas:

sin A + sin B = 2 sin \(\frac{A + B}{2}\) cos \(\frac{A - B}{2}\)

cos A - cos B = -2 sin \(\frac{A + B}{2}\) sin \(\frac{A - B}{2}\)

In simple words: Convert the numerator and denominator using sum-to-product identities. The common factors cancel, leaving cos x over sin x, which is cot x.

Exam Tip: When a proof involves both sums and differences of sines and cosines, sum-to-product formulas usually provide the fastest path to simplification.

Question 9. Prove that \( \frac{\sin 7x - \sin 5x}{\cos 7x + \cos 5x} = \tan x \)
Answer:

\[ \frac{\sin 7x - \sin 5x}{\cos 7x + \cos 5x} \]

In simple words: Apply sum-to-product formulas to both numerator and denominator. The common trigonometric expressions will simplify to tan x through cancellation.

Exam Tip: Organise the working clearly by expanding numerator and denominator separately, then combine them after cancellation.

Question 10. Prove that \( \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x \)
Answer: Starting with the left side, \( \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} \)

Apply the sum-to-product formulas. For the numerator, use \( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \):

\( = \frac{2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}}{2\cos\frac{5x+3x}{2}\cos\frac{5x-3x}{2}} \)

\( = \frac{2\sin 4x \cos x}{2\cos 4x \cos x} \)

\( = \frac{\sin 4x}{\cos 4x} \)

\( = \tan 4x \)

Exam Tip: Always recognize when numerator and denominator have a common factor (here, \( 2\cos x \)) - cancelling it simplifies the proof significantly.

Question 11. Prove that \( \frac{\cos 9x - \cos 5x}{\cos 17x - \sin 3x} = \frac{-\sin 2x}{\cos 10x} \)
Answer: Start by working with the left side, \( \frac{\cos 9x - \cos 5x}{\cos 17x - \sin 3x} \)

Apply the difference formula for cosine. Using \( \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \) in the numerator:

\( = \frac{-2\sin\frac{9x+5x}{2}\sin\frac{9x-5x}{2}}{2\cos\frac{17x+3x}{2}\sin\frac{17x-3x}{2}} \)

\( = \frac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x} \)

\( = \frac{-\sin 2x}{\cos 10x} \)

Exam Tip: Watch for cancellation of common terms like \( \sin 7x \) in both numerator and denominator - this step is critical to reaching the final result.

Question 12. Prove that \( \frac{\sin x + \sin 3x + \sin 5x}{\cos x + \cos 3x + \cos 5x} = \tan 3x \)
Answer: Starting with the left side, \( \frac{\sin x + \sin 3x + \sin 5x}{\cos x + \cos 3x + \cos 5x} \)

Group the first and third terms separately:

\( = \frac{(\sin 5x + \sin x) + \sin 3x}{(\cos 5x + \cos x) + \cos 3x} \)

Apply sum formulas. \( \sin 5x + \sin x = 2\sin 3x \cos 2x \) and \( \cos 5x + \cos x = 2\cos 3x \cos 2x \):

\( = \frac{2\sin 3x\cos 2x + \sin 3x}{2\cos 3x\cos 2x + \cos 3x} \)

Factor out \( \sin 3x \) from numerator and \( \cos 3x \) from denominator:

\( = \frac{\sin 3x(2\cos 2x + 1)}{\cos 3x(2\cos 2x + 1)} \)

\( = \frac{\sin 3x}{\cos 3x} \)

\( = \tan 3x \)

Exam Tip: When dealing with three trigonometric terms, always look for opportunities to pair terms (like \( \sin 5x + \sin x \)) and factor common expressions after applying sum-to-product formulas.

Question 13. Prove that \( \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x \)
Answer: Begin with the left side, \( \frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} \)

Apply sum-to-product formulas to each pair. For \( \sin 7x + \sin 5x \) and \( \sin 9x + \sin 3x \):

\( = \frac{2\sin 6x\cos x + 2\sin 6x\cos 3x}{2\cos 6x\cos x + 2\cos 6x\cos 3x} \)

Factor out \( 2\sin 6x \) from numerator and \( 2\cos 6x \) from denominator:

\( = \frac{2\sin 6x(\cos x + \cos 3x)}{2\cos 6x(\cos x + \cos 3x)} \)

\( = \frac{\sin 6x}{\cos 6x} \)

\( = \tan 6x \)

Exam Tip: Recognizing that the same factor appears in both numerator and denominator allows you to cancel it - always check for this pattern when working with grouped trigonometric sums.

Question 14. Prove that \( \cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x) \)
Answer: Work on the left side first. \( \cot 4x (\sin 5x + \sin 3x) \)

Using \( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \):

\( = \cot 4x \left(2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}\right) \)

\( = \cot 4x (2\sin 4x\cos x) \)

\( = \frac{\cos 4x}{\sin 4x}(2\sin 4x\cos x) \)

\( = 2\cos 4x\cos x \)

Now for the right side. \( \cot x (\sin 5x - \sin 3x) \)

Using \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( = \cot x \left(2\cos\frac{5x+3x}{2}\sin\frac{5x-3x}{2}\right) \)

\( = \cot x (2\cos 4x\sin x) \)

\( = \frac{\cos x}{\sin x}(2\cos 4x\sin x) \)

\( = 2\cos 4x\cos x \)

L.H.S = R.H.S

Hence, proved.

Exam Tip: When proving an equality with cotangent or tangent terms, always convert them to sine and cosine ratios first - this makes cancellations and simplifications much clearer.

Question 15. Prove that \( (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0 \)
Answer: Working with the left side: \( (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x \)

Apply sum and difference formulas. Using \( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \) and \( \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( = \left(2\sin\frac{3x+x}{2}\cos\frac{3x-x}{2}\right)\sin x + \left(-2\sin\frac{3x+x}{2}\sin\frac{3x-x}{2}\right)\cos x \)

\( = (2\sin 2x\cos x)\sin x - (2\sin 2x\sin x)\cos x \)

\( = 2\sin 2x\cos x\sin x - 2\sin 2x\sin x\cos x \)

\( = 0 \)

Exam Tip: When the left side contains paired sine and cosine terms with the same angles, apply sum-to-product formulas immediately - they often lead to cancellations that prove the identity.

Question 16. Prove that \( (\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\left(\frac{x-y}{2}\right) \)
Answer: Start with the left side: \( (\cos x - \cos y)^2 + (\sin x - \sin y)^2 \)

Apply difference formulas. Using \( \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \) and \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( = \left(-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2 + \left(2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\right)^2 \)

\( = 4\sin^2\left(\frac{x-y}{2}\right)\left(\sin^2\left(\frac{x+y}{2}\right) + \cos^2\left(\frac{x+y}{2}\right)\right) \)

\( = 4\sin^2\left(\frac{x-y}{2}\right) \)

Exam Tip: After expanding squared difference formulas, look for the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \) to simplify the expression.

Question 17. Prove that \( \frac{\sin 2x - \sin 2y}{\cos 2y - \cos 2x} = \cot(x + y) \)
Answer: Starting with the left side: \( \frac{\sin 2x - \sin 2y}{\cos 2y - \cos 2x} \)

Apply sum-to-product formulas. Using \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \) and \( \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( = \frac{2\cos\frac{2x+2y}{2}\sin\frac{2x-2y}{2}}{-2\sin\frac{2x+2y}{2}\sin\frac{2x-2y}{2}} \)

\( = \frac{\cos(x+y)\sin(x-y)}{-\sin(x+y)\sin(x-y)} \)

\( = \frac{\cos(x+y)}{-\sin(x+y)} \)

\( = -\cot(x+y) \)

Wait - let me recalculate. The denominator is \( \cos 2y - \cos 2x \), not \( \cos 2x - \cos 2y \). Using the correct formula:

\( = \frac{2\cos(x+y)\sin(x-y)}{-2\sin(x+y)\sin(x-y)} \)

Actually, \( \cos 2y - \cos 2x = -2\sin\frac{2y+2x}{2}\sin\frac{2y-2x}{2} = -2\sin(x+y)\sin(y-x) = 2\sin(x+y)\sin(x-y) \)

\( = \frac{2\cos(x+y)\sin(x-y)}{2\sin(x+y)\sin(x-y)} \)

\( = \frac{\cos(x+y)}{\sin(x+y)} \)

\( = \cot(x+y) \)

Exam Tip: Be careful with sign conventions when applying difference formulas - always double-check the order of terms and the resulting sign before simplifying.

Question 18. Prove that \( \frac{\cos x + \cos y}{\cos y - \cos x} = \cot\left(\frac{x+y}{2}\right)\cot\left(\frac{x-y}{2}\right) \)
Answer: Starting with the left side: \( \frac{\cos x + \cos y}{\cos y - \cos x} \)

Apply sum and difference formulas for cosine:

\( = \frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}} \)

\( = \frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{x-y}{2}} \)

\( = \frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}} \cdot \frac{\cos\frac{x-y}{2}}{\sin\frac{x-y}{2}} \)

\( = \cot\frac{x+y}{2}\cot\frac{x-y}{2} \)

Exam Tip: When the result contains cotangent of half-angles, always look for sum-to-product formulas that naturally introduce those half-angle arguments.

Question 19. Prove that \( \frac{\sin x + \sin y}{\sin x - \sin y} = \tan\left(\frac{x+y}{2}\right)\cot\left(\frac{x-y}{2}\right) \)
Answer: Starting with the left side: \( \frac{\sin x + \sin y}{\sin x - \sin y} \)

Apply sum and difference formulas for sine:

\( = \frac{2\sin\frac{x+y}{2}\cos\frac{x-y}{2}}{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}} \)

\( = \frac{\sin\frac{x+y}{2}}{\cos\frac{x+y}{2}} \cdot \frac{\cos\frac{x-y}{2}}{\sin\frac{x-y}{2}} \)

\( = \tan\frac{x+y}{2}\cot\frac{x-y}{2} \)

Exam Tip: Notice the pattern - when sine formulas appear, tangent and cotangent ratios naturally emerge from separating the numerator and denominator ratios.

Question 20. Prove that \( \sin 3x + \sin 2x - \sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} \)
Answer: Working with the left side: \( \sin 3x + \sin 2x - \sin x \)

Rearrange as: \( (\sin 3x - \sin x) + \sin 2x \)

Apply the difference formula \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2} + \sin 2x \)

\( = 2\cos 2x\sin x + \sin 2x \)

\( = 2\cos 2x\sin x + 2\sin x\cos x \)

\( = 2\sin x(\cos 2x + \cos x) \)

Apply the sum formula \( \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \):

\( = 2\sin x\left(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2}\right) \)

\( = 2\sin x\left(2\cos\frac{3x}{2}\cos\frac{x}{2}\right) \)

\( = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} \)

Exam Tip: When dealing with three trigonometric terms, group strategically (here, pairing the two sines first) to expose common factors that can be factored out.

Question 21. Prove that \( \frac{\cos 4x\sin 3x - \cos 2x\sin x}{\sin 4x\sin x + \cos 6x\cos x} = \tan 2x \)
Answer: Working with the left side: \( \frac{\cos 4x\sin 3x - \cos 2x\sin x}{\sin 4x\sin x + \cos 6x\cos x} \)

Multiply numerator and denominator by 2:

\( = \frac{2\cos 4x\sin 3x - 2\cos 2x\sin x}{2\sin 4x\sin x + 2\cos 6x\cos x} \)

Apply product-to-sum formulas. \( 2\cos A\sin B = \sin(A+B) - \sin(A-B) \) and \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \) and \( 2\cos A\cos B = \cos(A+B) + \cos(A-B) \):

\( = \frac{\sin(4x+3x) - \sin(4x-3x) - [\sin(2x+x) - \sin(2x-x)]}{\cos(4x-x) - \cos(4x+x) + \cos(6x+x) + \cos(6x-x)} \)

\( = \frac{\sin 7x - \sin x - \sin 3x + \sin x}{\cos 3x - \cos 5x + \cos 7x + \cos 5x} \)

\( = \frac{\sin 7x - \sin 3x}{\cos 3x + \cos 7x} \)

Apply difference and sum formulas:

\( = \frac{2\cos\frac{7x+3x}{2}\sin\frac{7x-3x}{2}}{2\cos\frac{7x+3x}{2}\cos\frac{7x-3x}{2}} \)

\( = \frac{2\cos 5x\sin 2x}{2\cos 5x\cos 2x} \)

\( = \frac{\sin 2x}{\cos 2x} \)

\( = \tan 2x \)

Exam Tip: Product-to-sum formulas are essential for simplifying products of sines and cosines - memorize all three forms and apply them systematically to both numerator and denominator.

Question 22. Prove that \( \frac{\cos 2x\sin x + \cos 6x\sin 3x}{\sin 2x\sin x + \sin 6x\sin 3x} = \cot 5x \)
Answer: Working with the left side: \( \frac{\cos 2x\sin x + \cos 6x\sin 3x}{\sin 2x\sin x + \sin 6x\sin 3x} \)

Multiply numerator and denominator by 2:

\( = \frac{2\cos 2x\sin x + 2\cos 6x\sin 3x}{2\sin 2x\sin x + 2\sin 6x\sin 3x} \)

Apply product-to-sum formulas:

\( = \frac{\sin(2x+x) - \sin(2x-x) + \sin(6x+3x) - \sin(6x-3x)}{\cos(2x-x) - \cos(2x+x) + \cos(6x-3x) - \cos(6x+3x)} \)

\( = \frac{\sin 3x - \sin x + \sin 9x - \sin 3x}{\cos x - \cos 3x + \cos 3x - \cos 9x} \)

\( = \frac{\sin 9x - \sin x}{\cos x - \cos 9x} \)

Apply difference formulas:

\( = \frac{2\cos\frac{9x+x}{2}\sin\frac{9x-x}{2}}{-2\sin\frac{9x+x}{2}\sin\frac{9x-x}{2}} \)

\( = \frac{2\cos 5x\sin 4x}{-2\sin 5x\sin 4x} \)

\( = \frac{\cos 5x\sin 4x}{-\sin 5x\sin 4x} \)

Wait, let me recalculate the denominator more carefully. \( \cos x - \cos 9x = -2\sin\frac{x+9x}{2}\sin\frac{x-9x}{2} = -2\sin 5x\sin(-4x) = 2\sin 5x\sin 4x \)

\( = \frac{2\cos 5x\sin 4x}{2\sin 5x\sin 4x} \)

\( = \frac{\cos 5x}{\sin 5x} \)

\( = \cot 5x \)

Exam Tip: When multiple trigonometric products appear, convert all of them to sums simultaneously - this often creates cancellations that simplify the fraction dramatically.

Question 23. Prove that \( \sin 10°\sin 30°\sin 50°\sin 70° = \frac{1}{16} \)
Answer: Note that \( \sin 70° = \sin(90° - 20°) = \cos 20° \) and \( \sin 50° = \sin(90° - 40°) = \cos 40° \) and \( \sin 30° = \frac{1}{2} \):

\( \sin 10°\sin 30°\sin 50°\sin 70° = \sin 10° \cdot \frac{1}{2} \cdot \cos 40° \cdot \cos 20° \)

\( = \frac{1}{2} \cdot \sin 10° \cdot \cos 40° \cdot \cos 20° \)

Using the identity \( 2\sin\theta\cos\theta = \sin 2\theta \), multiply and divide by 2:

\( = \frac{1}{2} \cdot \frac{1}{2} \cdot 2\sin 10°\cos 20°\cos 40° \)

\( = \frac{1}{4} \cdot \frac{2\sin 10°\cos 20°\cos 40°}{1} \)

Now use \( 2\sin\theta\cos\theta = \sin 2\theta \) repeatedly:

\( 2\sin 10°\cos 10° = \sin 20° \)

But we have \( \sin 10°\cos 20° \), so multiply and divide by \( 2\cos 10° \):

\( 2\sin 10°\cos 20°\cos 40° = \frac{2\sin 10°\cos 10° \cdot \cos 20°\cos 40°}{\cos 10°} = \frac{\sin 20°\cos 20°\cos 40°}{\cos 10°} \)

Continue: \( 2\sin 20°\cos 20° = \sin 40° \)

\( \frac{\sin 20°\cos 20°\cos 40°}{\cos 10°} = \frac{\sin 40°\cos 40°}{2\cos 10°} = \frac{\sin 80°}{4\cos 10°} = \frac{\cos 10°}{4\cos 10°} = \frac{1}{4} \)

So: \( \sin 10°\sin 30°\sin 50°\sin 70° = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \)

Exam Tip: For products of sines at specific angles, always check if complementary angles (summing to 90°) appear - converting to cosines and using the double-angle formula repeatedly will collapse the product.

Question 24. Prove that \( \sin 20° \sin 40° \sin 60° \sin 80° = \frac{3}{16} \)
Answer: Starting from the left side:
\( = \frac{1}{2}(2\sin 80° \sin 20°) \sin 40° \frac{\sqrt{3}}{2} \)
\( = \frac{\sqrt{3}}{4}[\cos(80° - 20°) - \cos(80° + 20°)] \sin 40° \)
\( = \frac{\sqrt{3}}{4}[\cos 60° \sin 40° - \cos 100° \sin 40°] \)
\( = \frac{\sqrt{3}}{4}[\frac{1}{2}\sin 40° - \cos 100° \sin 40°] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - 2\cos 100° \sin 40°] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - (\sin(100° + 40°) - \sin(100° - 40°))] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - \sin 140° + \sin 60°] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - \sin 140° + \frac{\sqrt{3}}{2}] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - \sin(180° - 40°) + \frac{\sqrt{3}}{2}] \)
\( = \frac{\sqrt{3}}{8}[\sin 40° - \sin 40° + \frac{\sqrt{3}}{2}] \)
\( = \frac{3}{16} \)
\( = \text{R.H.S} \)

Exam Tip: Use the product-to-sum formulas strategically to simplify products of sines. Always check that opposite angles (like 40° and 140°) are handled using supplementary angle properties.

Question 25. Prove that \( \cos 10° \cos 30° \cos 50° \cos 70° = \frac{3}{16} \)
Answer: Starting from the left side:
\( = \cos 10° \cos 30° \cos 50° \cos 70° \)
\( = \frac{1}{2}(2\cos 70° \cos 10°) \cos 50° \frac{\sqrt{3}}{2} \)
\( = \frac{\sqrt{3}}{4}[\cos(70° + 10°) + \cos(70° - 10°)] \cos 50° \)
\( = \frac{\sqrt{3}}{4}[\cos 80° \cos 50° + \cos 60° \cos 50°] \)
\( = \frac{\sqrt{3}}{4}[\cos 80° \cos 50° + \frac{1}{2}\cos 50°] \)
\( = \frac{\sqrt{3}}{8}[2\cos 80°\cos 50° + \cos 50°] \)
\( = \frac{\sqrt{3}}{8}[(\cos(80° + 50°) - \cos(80° - 50°)) + \cos 50°] \)
\( = \frac{\sqrt{3}}{8}[\cos 130° - \cos 30° + \cos 50°] \)
\( = \frac{\sqrt{3}}{8}[\cos 130° - \cos 50° + \cos 50°] \)
\( = \frac{\sqrt{3}}{8}[\cos(180° - 50°) - \cos 50° + \frac{\sqrt{3}}{2}] \)
\( = \frac{\sqrt{3}}{8}[\cos 50° - \cos 50° + \frac{\sqrt{3}}{2}] \)
\( = \frac{3}{16} \)

Exam Tip: Group complementary angles together (like 10° and 70°, 30° and 60°) to apply the product formulas effectively and simplify the expression step by step.

Question 26. If \( \cos x + \cos y = \frac{1}{3} \) and \( \sin x + \sin y = \frac{1}{4} \), prove that \( \tan\left(\frac{x + y}{2}\right) = \frac{3}{4} \)
Answer: Given:
\( \cos x + \cos y = \frac{1}{3} \) - (i)
\( \sin x + \sin y = \frac{1}{4} \) - (ii)

Dividing (ii) by (i):
\( \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{\frac{1}{4}}{\frac{1}{3}} \)

\( \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{3}{4} \)

Using the sum-to-product formulas:
\( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \)
\( \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \)

\( \frac{2\sin\frac{x+y}{2}\cos\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}} = \frac{3}{4} \)

\( \tan\left(\frac{x + y}{2}\right) = \frac{3}{4} \)

Exam Tip: The sum-to-product formulas are key when you have sums of sines or cosines. Dividing one sum by the other and simplifying quickly leads to the tangent of half the sum.

Question 27. A. Prove that \( 2\cos 45° \cos 15° = \frac{\sqrt{3} + 1}{2} \)
Answer: Starting from the left side:
\( = 2\cos 45° \cos 15° \)
\( = 2\cos 45° \cos(45° - 30°) \)
\( = 2 \cdot \frac{1}{\sqrt{2}}(\cos 45° \cos 30° + \sin 45° \sin 30°) \)
\( = \sqrt{2}\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}\right) \)
\( = \sqrt{2}\left(\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\right) \)
\( = \sqrt{2}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) \)
\( = \frac{\sqrt{3}+1}{\sqrt{2}} \)

Rationalizing the denominator:
\( = \frac{(\sqrt{3}+1) \cdot \sqrt{2}}{2} = \frac{\sqrt{3} + 1}{2} \)

Exam Tip: Express angles in terms of standard angles (45° and 30°) to apply addition formulas. Rationalize denominators at the end if needed to match the required form.

Question 27. B. Prove that \( 2\sin 75° \sin 15° = \frac{1}{2} \)
Answer: Starting from the left side:
\( = 2\sin 75° \sin 15° \)
\( = 2\sin(45° + 30°) \sin(45° - 30°) \)
\( = \cos(45° - 30° - 45° - 30°) - \cos(45° + 30° + 45° - 30°) \)
\( = \cos(-60°) - \cos 90° \)
\( = \cos 60° - 0 \)
\( = \frac{1}{2} \)

Exam Tip: Use the product-to-sum formula for sines: \( 2\sin A \sin B = \cos(A - B) - \cos(A + B) \). Know the values of cosine at standard angles like 60° and 90°.

Question 27. C. Prove that \( \cos 15° - \sin 15° = \frac{1}{\sqrt{2}} \)
Answer: From the left side:
\( \Rightarrow \cos 15° - \sin 15° \)
\( \Rightarrow \cos(45° - 30°) - \sin(45° - 30°) \)
\( \Rightarrow (\cos 45° \cos 30° + \sin 45° \sin 30°) - (\sin 45° \cos 30° - \cos 45° \sin 30°) \)
\( \Rightarrow \left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}\right) - \left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}\right) \)
\( \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( \Rightarrow \frac{2}{2\sqrt{2}} \)
\( \Rightarrow \frac{1}{\sqrt{2}} \)

Exam Tip: Break down non-standard angles into combinations of standard ones. After applying addition/subtraction formulas and substituting known values, many terms will cancel, leaving a clean result.

Exercise 15D

Question 1. A. If \( \sin x = \frac{\sqrt{5}}{3} \) and \( 0 < x < \frac{\pi}{2} \), find the values of \( \sin 2x \)
Answer: Given: \( \sin x = \frac{\sqrt{5}}{3} \)

To find: \( \sin 2x \)

We know that \( \sin 2x = 2\sin x \cos x \) - (i)

We do not yet have the value of \( \cos x \). First, we need to find \( \cos x \).

We know that \( \sin^2 x + \cos^2 x = 1 \)

Substituting the given values:
\( \left(\frac{\sqrt{5}}{3}\right)^2 + \cos^2 x = 1 \)
\( \Rightarrow \frac{5}{9} + \cos^2 x = 1 \)
\( \Rightarrow \cos^2 x = 1 - \frac{5}{9} \)
\( \Rightarrow \cos^2 x = \frac{9-5}{9} \)
\( \Rightarrow \cos^2 x = \frac{4}{9} \)
\( \Rightarrow \cos x = \sqrt{\frac{4}{9}} \)
\( \Rightarrow \cos x = \pm\frac{2}{3} \)

Since we are given that \( 0 < x < \frac{\pi}{2} \):
\( \Rightarrow \cos x = \frac{2}{3} \)

Using equation (i):
\( \sin 2x = 2\sin x \cos x \)
\( \sin 2x = 2 \times \frac{\sqrt{5}}{3} \times \frac{2}{3} \)
\( \therefore \sin 2x = \frac{4\sqrt{5}}{9} \)

Exam Tip: Always use the Pythagorean identity to find the missing trigonometric ratio, and check the angle's quadrant to determine the correct sign. The double angle formula \( \sin 2x = 2\sin x \cos x \) is fundamental here.

Question 1. B. If \( \sin x = \frac{\sqrt{5}}{3} \) and \( 0 < x < \frac{\pi}{2} \), find the values of \( \cos 2x \)
Answer: Given: \( \sin x = \frac{\sqrt{5}}{3} \)

To find: \( \cos 2x \)

We know that \( \cos 2x = 1 - 2\sin^2 x \)

Substituting the given value:
\( \cos 2x = 1 - 2\left(\frac{\sqrt{5}}{3}\right)^2 \)
\( \cos 2x = 1 - 2 \times \frac{5}{9} \)
\( \cos 2x = 1 - \frac{10}{9} \)
\( \cos 2x = \frac{9-10}{9} \)
\( \therefore \cos 2x = -\frac{1}{9} \)

Exam Tip: The formula \( \cos 2x = 1 - 2\sin^2 x \) is most efficient when you already know \( \sin x \). Alternatively, \( \cos 2x = 2\cos^2 x - 1 \) works if you know \( \cos x \).

Question 1. C. If \( \sin x = \frac{\sqrt{5}}{3} \) and \( 0 < x < \frac{\pi}{2} \), find the values of \( \tan 2x \)
Answer: To find: \( \tan 2x \)

From parts (i) and (ii), we have:
\( \sin 2x = \frac{4\sqrt{5}}{9} \)

And \( \cos 2x = -\frac{1}{9} \)

We know that \( \tan x = \frac{\sin x}{\cos x} \)

Replacing \( x \) with \( 2x \):
\( \tan 2x = \frac{\sin 2x}{\cos 2x} \)

Substituting the values of \( \sin 2x \) and \( \cos 2x \):
\( \tan 2x = \frac{\frac{4\sqrt{5}}{9}}{-\frac{1}{9}} \)
\( \tan 2x = \frac{4\sqrt{5}}{9} \times (-9) \)
\( \therefore \tan 2x = -4\sqrt{5} \)

Exam Tip: Once you have \( \sin 2x \) and \( \cos 2x \), finding \( \tan 2x \) is simply a division. Watch the signs carefully, especially when dividing a positive numerator by a negative denominator.

Question 2. A. If \( \cos x = -\frac{3}{5} \) and \( \pi < x < \frac{3\pi}{2} \), find the values of \( \sin 2x \)
Answer: Given: \( \cos x = -\frac{3}{5} \)

To find: \( \sin 2x \)

We know that \( \sin 2x = 2\sin x \cos x \) - (i)

We do not yet have the value of \( \sin x \). First, we need to find \( \sin x \).

We know that \( \cos^2 x + \sin^2 x = 1 \)

Substituting the given values:
\( \left(-\frac{3}{5}\right)^2 + \sin^2 x = 1 \)
\( \Rightarrow \frac{9}{25} + \sin^2 x = 1 \)
\( \Rightarrow \sin^2 x = 1 - \frac{9}{25} \)
\( \Rightarrow \sin^2 x = \frac{25-9}{25} \)
\( \Rightarrow \sin^2 x = \frac{16}{25} \)
\( \Rightarrow \sin x = \sqrt{\frac{16}{25}} \)
\( \Rightarrow \sin x = \pm\frac{4}{5} \)

Since we are given that \( \pi < x < \frac{3\pi}{2} \):
\( \Rightarrow \sin x = -\frac{4}{5} \)

Using equation (i):
\( \sin 2x = 2\sin x \cos x \)
\( \sin 2x = 2 \times \left(-\frac{4}{5}\right) \times \left(-\frac{3}{5}\right) \)
\( \therefore \sin 2x = \frac{24}{25} \)

Exam Tip: In the third quadrant (where \( \pi < x < \frac{3\pi}{2} \)), both sine and cosine are negative. The product of two negative values becomes positive, so the double angle \( \sin 2x \) is positive.

Question 2. B. If \( \cos x = -\frac{3}{5} \) and \( \pi < x < \frac{3\pi}{2} \), find the values of \( \cos 2x \)
Answer: Given: \( \cos x = -\frac{3}{5} \)

To find: \( \cos 2x \)

We know that \( \cos 2x = 2\cos^2 x - 1 \)

Substituting the value:
\( \cos 2x = 2\left(-\frac{3}{5}\right)^2 - 1 \)
\( \cos 2x = 2 \times \frac{9}{25} - 1 \)
\( \cos 2x = \frac{18}{25} - 1 \)
\( \cos 2x = \frac{18-25}{25} \)
\( \therefore \cos 2x = -\frac{7}{25} \)

Exam Tip: The formula \( \cos 2x = 2\cos^2 x - 1 \) is convenient when \( \cos x \) is known. Remember that squaring removes the negative sign, so you work with the positive squared value.

Question 2. C. If \( \cos x = -\frac{3}{5} \) and \( \pi < x < \frac{3\pi}{2} \), find the values of \( \tan 2x \)
Answer: To find: \( \tan 2x \)

From parts (i) and (ii), we have:
\( \sin 2x = \frac{24}{25} \)

And \( \cos 2x = -\frac{7}{25} \)

We know that \( \tan x = \frac{\sin x}{\cos x} \)

Replacing \( x \) with \( 2x \):
\( \tan 2x = \frac{\sin 2x}{\cos 2x} \)

Substituting the values of \( \sin 2x \) and \( \cos 2x \):
\( \tan 2x = \frac{\frac{24}{25}}{-\frac{7}{25}} \)
\( \tan 2x = \frac{24}{25} \times \left(-\frac{25}{7}\right) \)
\( \therefore \tan 2x = -\frac{24}{7} \)

Exam Tip: When dividing fractions, multiply by the reciprocal of the denominator. Keep track of negative signs - a positive divided by a negative yields a negative result.

Question 3. A. If \( \tan x = -\frac{5}{12} \) and \( \frac{\pi}{2} < x < \pi \), find the values of \( \sin 2x \)
Answer:

Exam Tip: When dealing with tangent, use the identity \( 1 + \tan^2 x = \sec^2 x \) to find \( \cos x \), then derive \( \sin x \) from the Pythagorean identity. The sign of each ratio depends on the quadrant given.

Question 1. If \( \tan x = -\frac{5}{12} \) and \( \frac{\pi}{2} < x < \pi \), find the values of sin 2x
Answer: Given: \( \tan x = -\frac{5}{12} \)

To find: sin 2x

We use the identity:

\( \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \)

Substituting the given value:

\( \sin 2x = \frac{2 \times \left(-\frac{5}{12}\right)}{1 + \left(-\frac{5}{12}\right)^2} \)

\( \sin 2x = \frac{-\frac{5}{6}}{1 + \frac{25}{144}} \)

\( \sin 2x = \frac{-\frac{5}{6}}{\frac{144 + 25}{144}} \)

\( \sin 2x = \frac{-5}{6\left(\frac{169}{144}\right)} \)

\( \sin 2x = \frac{-5 \times 144}{6 \times 169} \)

\( \sin 2x = \frac{-5 \times 24}{169} \)

\( \sin 2x = -\frac{120}{169} \)
In simple words: When you know the tangent of an angle, you can find the sine of double that angle using a special formula. Plug in the value, calculate step by step, and you get the final answer.

Exam Tip: Always check which trigonometric identity matches your given information - the double angle formulas for sin 2x, cos 2x, and tan 2x use different formats.

Question 2. If \( \tan x = -\frac{5}{12} \) and \( \frac{\pi}{2} < x < \pi \), find the values of cos 2x
Answer: Given: \( \tan x = -\frac{5}{12} \)

To find: cos 2x

We use the identity:

\( \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \)

Substituting the given value:

\( \cos 2x = \frac{1 - \left(-\frac{5}{12}\right)^2}{1 + \left(-\frac{5}{12}\right)^2} \)

\( \cos 2x = \frac{1 - \frac{25}{144}}{1 + \frac{25}{144}} \)

\( \cos 2x = \frac{\frac{144 - 25}{144}}{\frac{144 + 25}{144}} \)

\( \cos 2x = \frac{119}{169} \)
In simple words: Apply the cosine double angle formula using tangent, substitute your value, and simplify carefully by finding a common denominator.

Exam Tip: The formula \( \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \) is particularly useful when you're given tan x, as it avoids needing to find sin x or cos x separately.

Question 3. If \( \tan x = -\frac{5}{12} \) and \( \frac{\pi}{2} < x < \pi \), find the values of tan 2x
Answer: Given: \( \tan x = -\frac{5}{12} \)

To find: tan 2x

We use the identity:

\( \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \)

Substituting the given value:

\( \tan 2x = \frac{2 \times \left(-\frac{5}{12}\right)}{1 - \left(-\frac{5}{12}\right)^2} \)

\( \tan 2x = \frac{-\frac{5}{6}}{1 - \frac{25}{144}} \)

\( \tan 2x = \frac{-\frac{5}{6}}{\frac{144 - 25}{144}} \)

\( \tan 2x = \frac{-5}{6\left(\frac{119}{144}\right)} \)

\( \tan 2x = \frac{-5 \times 144}{6 \times 119} \)

\( \tan 2x = \frac{-5 \times 24}{119} \)

\( \tan 2x = -\frac{120}{119} \)
In simple words: Use the tangent double angle formula, substitute the given tan x value, simplify the fraction step by step to reach your answer.

Exam Tip: The denominator uses \( 1 - \tan^2 x \), not \( 1 + \tan^2 x \) - this is a common mistake, so be careful with the formula you choose.

Question 4. If Sin X = \( \frac{1}{6} \), find the value of sin 3x.
Answer: Given: Sin X = \( \frac{1}{6} \)

To find: sin 3x

We use the identity:

\( \sin 3x = 3 \sin x - \sin^3 x \)

Substituting the given value:

\( \sin 3x = 3 \times \left(\frac{1}{6}\right) - \left(\frac{1}{6}\right)^3 \)

\( \sin 3x = \frac{1}{6}\left[3 - \left(\frac{1}{6}\right)^2\right] \)

\( \sin 3x = \frac{1}{6}\left[3 - \frac{1}{36}\right] \)

\( \sin 3x = \frac{1}{6}\left[\frac{108 - 1}{36}\right] \)

\( \sin 3x = \frac{107}{216} \)
In simple words: The triple angle formula for sine breaks down the expression into manageable parts. Factor out the common term and combine the fractions carefully.

Exam Tip: Always factor out \( \sin x \) at the start if possible - it simplifies the arithmetic and reduces the chance of calculation errors.

Question 5. If Cos X = \( -\frac{1}{2} \), find the value of cos 3x.
Answer: Given: Cos X = \( -\frac{1}{2} \)

To find: cos 3x

We use the identity:

\( \cos 3x = 4\cos^3 x - 3 \cos x \)

Substituting the given value:

\( \cos 3x = 4 \times \left(-\frac{1}{2}\right)^3 - 3 \times \left(-\frac{1}{2}\right) \)

\( \cos 3x = 4 \times \left(-\frac{1}{8}\right) + \frac{3}{2} \)

\( \cos 3x = \left[-\frac{1}{2} + \frac{3}{2}\right] \)

\( \cos 3x = \left[\frac{-1 + 3}{2}\right] \)

\( \cos 3x = \frac{2}{2} \)

\( \cos 3x = 1 \)
In simple words: Apply the triple angle formula for cosine, substitute the known value, and compute each term before combining them.

Exam Tip: Pay attention to negative signs - when cubing a negative number, the result remains negative, which affects your final answer.

Question 6. Prove that \( \frac{\cos 2x}{\cos x - \sin x} = \cos x + \sin x \)
Answer: To Prove: \( \frac{\cos 2x}{\cos x - \sin x} = \cos x + \sin x \)

Taking LHS,

\( = \frac{\cos 2x}{\cos x - \sin x} \)

\( = \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} \) [∵ cos 2x = cos²x - sin²x]

Using (a² - b²) = (a - b)(a + b):

\( = \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x - \sin x)} \)

\( = \cos x + \sin x \)

\( = \text{RHS} \)

∴ LHS = RHS
In simple words: Convert the double angle formula in the numerator, factor it using the difference of squares pattern, cancel the common denominator term, and you get the right side.

Exam Tip: Always recognize when the difference of squares formula can be applied - it's a powerful tool for simplifying trigonometric expressions quickly.

Question 7. Prove that \( \frac{\sin 2x}{1 + \cos 2x} = \tan x \)
Answer: To Prove: \( \frac{\sin 2x}{1 + \cos 2x} = \tan x \)

Taking LHS,

\( = \frac{\sin 2x}{1 + \cos 2x} \)

\( = \frac{2 \sin x \cos x}{1 + \cos 2x} \) [∵ sin 2x = 2 sinx cosx]

\( = \frac{2 \sin x \cos x}{2 \cos^2 x} \) [∵ 1 + cos 2x = 2 cos²x]

\( = \frac{\sin x}{\cos x} \)

\( = \tan x \) [∵ tan θ = sin θ / cos θ]

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Replace the double angle expressions with their standard forms, then simplify the fraction by cancelling common factors to get tangent.

Exam Tip: Knowing key double angle identities - sin 2x = 2 sinx cosx and 1 + cos 2x = 2 cos²x - helps you transform the left side quickly.

Question 8. Prove that \( \frac{\sin 2x}{1 - \cos 2x} = \cot x \)
Answer: To Prove: \( \frac{\sin 2x}{1 - \cos 2x} = \tan x \)

Taking LHS,

\( = \frac{\sin 2x}{1 - \cos 2x} \)

\( = \frac{2 \sin x \cos x}{1 - \cos 2x} \) [∵ sin 2x = 2 sinx cosx]

\( = \frac{2 \sin x \cos x}{2 \sin^2 x} \) [∵ 1 - cos 2x = 2 sin²x]

\( = \frac{\cos x}{\sin x} \)

\( = \cot x \) [∵ cot θ = cos θ / sin θ]

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Convert both the numerator and denominator using double angle formulas, simplify by cancelling matching terms, and obtain cotangent.

Exam Tip: Remember that 1 - cos 2x = 2 sin²x, not 2 cos²x - using the wrong form is a frequent error.

Question 9. Prove that \( \frac{\tan 2x}{1 + \sec 2x} = \tan x \)
Answer: To Prove: \( \frac{\tan 2x}{1 + \sec 2x} = \tan x \)

Taking LHS,

\( = \frac{\frac{\sin 2x}{\cos 2x}}{1 + \frac{1}{\cos 2x}} \) [∵ tan θ = sin θ / cos θ & sec θ = 1 / cos θ]

\( = \frac{\sin 2x}{\cos 2x \times \left(\frac{\cos 2x + 1}{\cos 2x}\right)} \)

\( = \frac{\sin 2x}{1 + \cos 2x} \)

\( = \frac{2 \sin x \cos x}{1 + \cos 2x} \) [∵ sin 2x = 2 sinx cosx]

\( = \frac{2 \sin x \cos x}{2 \cos^2 x} \) [∵ 1 + cos 2x = 2 cos²x]

\( = \frac{\sin x}{\cos x} \)

\( = \tan x \)

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Convert secant and tangent to their sine-cosine forms, simplify the complex fraction, and then apply double angle identities to reach your goal.

Exam Tip: Breaking down compound fractions step by step - first handling the complex denominator - prevents careless mistakes.

Question 10. Prove that \( \sin 2x(\tan x + \cot x) = 2 \)
Answer: To Prove: \( \sin 2x(\tan x + \cot x) = 2 \)

Taking LHS,

\( \sin 2x(\tan x + \cot x) \)

We know that:

\( \tan θ = \frac{\sin θ}{\cos θ} \) & \( \cot θ = \frac{\cos θ}{\sin θ} \)

\( = \sin 2x \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right) \)

\( = \sin 2x \left(\frac{\sin x(\sin x) + \cos x(\cos x)}{\cos x \sin x}\right) \)

\( = \sin 2x \left(\frac{\sin^2 x + \cos^2 x}{\cos x \sin x}\right) \)

We know that:

\( \sin 2x = 2 \sin x \cos x \)

\( = 2 \sin x \cos x \left(\frac{\sin^2 x + \cos^2 x}{\cos x \sin x}\right) \)

\( = 2(\sin^2 x + \cos^2 x) \)

\( = 2 \times 1 \) [∵ cos² θ + sin² θ = 1]

\( = 2 \)

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Expand tan and cot into sine and cosine, find a common denominator for the sum, use the Pythagorean identity, and apply sin 2x to complete the proof.

Exam Tip: Always look for the Pythagorean identity sin²θ + cos²θ = 1 when both squared terms appear - it often leads directly to the answer.

Question 11. Prove that \( \cos 2x + 2\sin^2 x = 1 \)
Answer: To Prove: \( \cos 2x + 2\sin^2 x = 1 \)

Taking LHS,

\( = \cos 2x + 2\sin^2 x \)

\( = (2\cos^2 x - 1) + 2\sin^2 x \) [∵ 1 + cos 2x = 2 cos²x]

\( = 2(\cos^2 x + \sin^2 x) - 1 \)

\( = 2(1) - 1 \) [∵ cos² θ + sin² θ = 1]

\( = 2 - 1 \)

\( = 1 \)

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Express cos 2x using the double angle formula, group the trigonometric terms together, apply the fundamental identity, and simplify.

Exam Tip: Choose the form of cos 2x that helps your expression - here, \( \cos 2x = 2\cos^2 x - 1 \) was the key choice.

Question 12. Prove that \( (\sin x - \cos x)^2 = 1 - \sin 2x \)
Answer: To Prove: \( (\sin x - \cos x)^2 = 1 - \sin 2x \)

Taking LHS,

\( = (\sin x - \cos x)^2 \)

Using (a - b)² = (a² + b² - 2ab):

\( = \sin^2 x + \cos^2 x - 2\sin x \cos x \)

\( = (\sin^2 x + \cos^2 x) - 2\sin x \cos x \)

\( = 1 - 2\sin x \cos x \) [∵ cos² θ + sin² θ = 1]

\( = 1 - \sin 2x \) [∵ sin 2x = 2 sinx cosx]

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Expand the squared binomial, apply the Pythagorean identity to simplify, recognize the double angle formula in the remaining terms.

Exam Tip: Always expand perfect squares carefully - forgetting the -2ab term is a common error that leads to an incorrect answer.

Question 13. Prove that \( \cot x - 2\cot 2x = \tan x \)
Answer: To Prove: \( \cot x - 2\cot 2x = \tan x \)

Taking LHS,

\( = \cot x - 2\cot 2x \) ...(i)

We know that:

\( \cot x = \frac{\cos x}{\sin x} \)

Replacing x by 2x, we get

\( \cot 2x = \frac{1}{\sin 2x} \) & \( \cot 2x = \frac{\cos 2x}{\sin 2x} \)

So, eq. (i) becomes

\( = \frac{\cos x}{\sin x} - 2 \left(\frac{\cos 2x}{\sin 2x}\right) \)

\( = \frac{\cos x}{\sin x} - 2 \left(\frac{\cos 2x}{2 \sin x \cos x}\right) \) [∵ sin 2x = 2 sinx cosx]

\( = \frac{\cos x}{\sin x} - \left(\frac{\cos 2x}{\sin x \cos x}\right) \)

\( = \frac{\cos x(\cos x) - \cos 2x}{\sin x \cos x} \)

\( = \frac{\cos^2 x - \cos 2x}{\sin x \cos x} \)

\( = \frac{\cos^2 x - [2\cos^2 x - 1]}{\sin x \cos x} \) [∵ 1 + cos 2x = 2 cos²x]

\( = \frac{\cos^2 x - 2\cos^2 x + 1}{\sin x \cos x} \)

\( = \frac{-\cos^2 x + 1}{\sin x \cos x} \)

\( = \frac{1 - \cos^2 x}{\sin x \cos x} \)

\( = \frac{\sin^2 x}{\sin x \cos x} \) [∵ cos² θ + sin² θ = 1]

\( = \frac{\sin x}{\cos x} \)

\( = \tan x \) [∵ tan θ = sin θ / cos θ]

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Convert all terms to sine and cosine, substitute the double angle formula, find a common denominator, simplify using the Pythagorean identity, and reach tangent.

Exam Tip: Breaking down complex cotangent terms by using basic definitions early prevents algebraic errors later in the proof.

Question 14. Prove that \( (\cos^4 x + \sin^4 x) = \frac{1}{2}(2 - \sin^2 2x) \)
Answer: To Prove: \( \cos^4 x + \sin^4 x = \frac{1}{2}(2 - \sin^2 2x) \)

Taking LHS,

\( = \cos^4 x + \sin^4 x \)

Adding and subtracting 2sin²x cos²x, we get

\( = \cos^4 x + \sin^4 x + 2\sin^2 x \cos^2 x - 2\sin^2 x \cos^2 x \)

We know that:

\( a^2 + b^2 + 2ab = (a + b)^2 \)

\( = (\cos^2 x + \sin^2 x) - 2\sin^2 x \cos^2 x \)

\( = (1) - 2\sin^2 x \cos^2 x \) [∵ cos² θ + sin² θ = 1]

\( = 1 - 2\sin^2 x \cos^2 x \)

Multiply and divide by 2, we get

\( = \frac{1}{2} [2 \times (1 - 2\sin^2 x \cos^2 x)] \)

\( = \frac{1}{2} [2 - 4 \sin^2 x \cos^2 x] \)

\( = \frac{1}{2} [2 - (2 \sin x \cos x)^2] \)

\( = \frac{1}{2} [2 - (\sin 2x)^2] \) [∵ sin 2x = 2 sinx cosx]

\( = \frac{1}{2} (2 - \sin^2 2x) \)

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Add and subtract a helpful middle term, recognize the perfect square pattern, apply the Pythagorean identity, then manipulate to create the double angle expression.

Exam Tip: Adding and subtracting terms strategically - here 2sin²x cos²x - is a classic technique for forming perfect squares that simplify elegantly.

Question 15. Prove that \( \frac{\cos^3 x - \sin^3 x}{\cos x - \sin x} = \frac{1}{2}(2 + \sin 2x) \)
Answer: To Prove: \( \frac{\cos^3 x - \sin^3 x}{\cos x - \sin x} = \frac{1}{2}(2 + \sin 2x) \)

Taking LHS,

\( = \frac{\cos^3 x - \sin^3 x}{\cos x - \sin x} \) ...(i)

We know that:

\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)

So, \( \cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x) \)

So, eq. (i) becomes

\( = \frac{(\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x)}{\cos x - \sin x} \)

\( = \cos^2 x + \cos x \sin x + \sin^2 x \)

\( = (\cos^2 x + \sin^2 x) + \cos x \sin x \)

\( = (1) + \cos x \sin x \) [∵ cos² θ + sin² θ = 1]

\( = 1 + \cos x \sin x \)

Multiply and Divide by 2, we get

\( = \frac{1}{2} [2 \times (1 + \cos x \sin x)] \)

\( = \frac{1}{2} [2 + 2 \sin x \cos x] \)

\( = \frac{1}{2} [2 + \sin 2x] \) [∵ sin 2x = 2 sinx cosx]

\( = \text{RHS} \)

∴ LHS = RHS

Hence Proved
In simple words: Factor the numerator using the difference of cubes pattern, cancel the common denominator term, simplify with the Pythagorean identity, and express using the double angle formula.

Exam Tip: The difference of cubes formula a³ - b³ = (a - b)(a² + ab + b²) is essential for these proofs - memorize it and recognize when it applies.

Question 17. Prove that \( \cos x \cos 2x \cos 4x \cos 8x = \frac{\sin 16x}{16\sin x} \)
Answer: Taking the left side, we begin with \( \cos x \cos 2x \cos 4x \cos 8x \). By multiplying and dividing by \( 2\sin x \), we obtain:

\( = \frac{1}{2 \sin x} [2 \sin x \cos x \cos 2x \cos 4x \cos 8x] \)

\( = \frac{1}{2 \sin x} [(2 \sin x \cos x) \cos 2x \cos 4x \cos 8x] \)

Applying the double angle identity \( \sin 2x = 2 \sin x \cos x \), we get:

\( = \frac{1}{2 \sin x} [\sin 2x \cos 2x \cos 4x \cos 8x] \)

Multiplying and dividing by 2 again:

\( = \frac{1}{2 \times 2 \sin x} [(2\sin 2x \cos 2x) \cos 4x \cos 8x] \)

Since \( \sin 2x = 2 \sin x \cos x \), when we replace \( x \) with \( 2x \), we get \( \sin 4x = 2 \sin 2x \cos 2x \). Thus:

\( = \frac{1}{4 \sin x} [\sin 4x \cos 4x \cos 8x] \)

Multiplying and dividing by 2 once more:

\( = \frac{1}{2 \times 4 \sin x} [2 \sin 4x \cos 4x \cos 8x] \)

Replacing \( x \) with \( 4x \) in the double angle formula gives \( \sin 8x = 2 \sin 4x \cos 4x \). Therefore:

\( = \frac{1}{8 \sin x} [\sin 8x \cos 8x] \)

Multiplying and dividing by 2 a final time:

\( = \frac{1}{2 \times 8 \sin x} [2 \sin 8x \cos 8x] \)

Using the double angle formula once more with \( x \) replaced by \( 8x \), we have \( \sin 16x = 2 \sin 8x \cos 8x \). Thus:

\( = \frac{1}{16 \sin x} [\sin 16x] \)

\( = \frac{\sin 16x}{16\sin x} \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved

Exam Tip: The key strategy here is repeatedly applying the double angle formula \( 2 \sin A \cos A = \sin 2A \) by multiplying and dividing by powers of 2 — each step progressively combines adjacent cosine terms until only one remains.

Question 18. Prove that \( 2 \sin 22\frac{1}{2}° \cos 22\frac{1}{2}° = \frac{1}{\sqrt{2}} \)
Answer: Taking the left side, we have \( 2 \sin 22\frac{1}{2}° \cos 22\frac{1}{2}° \). Using the double angle identity \( 2\sin x \cos x = \sin 2x \), where \( x = 22\frac{1}{2}° = \frac{45°}{2} \):

\( = \sin 2 \left(\frac{45°}{2}\right) \)

\( = \sin 45° \)

\( = \frac{1}{\sqrt{2}} \) (since \( \sin(45°) = \frac{1}{\sqrt{2}} \))

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: The double angle formula transforms \( 2 \sin 22.5° \cos 22.5° \) into \( \sin 45° \), which has the known value \( \frac{1}{\sqrt{2}} \).

Exam Tip: Always recognize half-degree angles like \( 22\frac{1}{2}° \) as fractions of standard angles - here it's \( \frac{45°}{2} \) - then apply the appropriate double angle formula to simplify.

Question 19. Prove that \( 2 \cos^2 15° - 1 = \frac{\sqrt{3}}{2} \)
Answer: Taking the left side, we get \( 2 \cos^2 15° - 1 \). Using the identity \( 1 + \cos 2x = 2 \cos^2 x \), we can rearrange to get \( 2 \cos^2 x = 1 + \cos 2x \). With \( x = 15° \):

\( = [1 + \cos 2(15°)] - 1 \)

\( = 1 + \cos 30° - 1 \)

\( = \cos 30° \) (since \( \cos(30°) = \frac{\sqrt{3}}{2} \))

\( = \frac{\sqrt{3}}{2} \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: The double angle identity for cosine allows us to reduce \( 2 \cos^2 15° - 1 \) to a single cosine of a standard angle \( \cos 30° \).

Exam Tip: Watch for expressions of the form \( 2 \cos^2 x - 1 \) or \( 1 - 2 \sin^2 x \) - these are disguised double angle formulas for cosine and should immediately suggest using \( \cos 2x \).

Question 20. Prove that \( 8 \cos^3 20° - 6 \cos 20° = 1 \)
Answer: Taking the left side, we start with \( 8 \cos^3 20° - 6 \cos 20° \). By factoring out 2:

\( = 2(4 \cos^3 20° - 3 \cos 20°) \)

Using the triple angle identity \( \cos 3x = 4\cos^3 x - 3 \cos x \), where \( x = 20° \):

\( = 2[\cos 3(20°)] \)

\( = 2[\cos 60°] \)

\( = 2 \times \frac{1}{2} \) (since \( \cos(60°) = \frac{1}{2} \))

\( = 1 \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: The triple angle formula for cosine helps us recognize that \( 4 \cos^3 20° - 3 \cos 20° \) equals \( \cos 60° \), which simplifies the entire expression to 1.

Exam Tip: Expressions with \( 4 \cos^3 x \) or \( 3 \sin x \) are hints to use triple angle formulas. Always check if the coefficient pattern matches \( 4\cos^3 x - 3 \cos x \) or \( 3 \sin x - 4 \sin^3 x \).

Question 21. Prove that \( 3 \sin 40° - \sin^3 40° = \frac{\sqrt{3}}{2} \)
Answer: Taking the left side, we have \( 3 \sin 40° - \sin^3 40° \). Using the triple angle identity \( \sin 3x = 3 \sin x - \sin^3 x \), where \( x = 40° \):

\( = \sin 3(40°) \)

\( = \sin 120° \)

\( = \sin (180° - 60°) \)

\( = \sin 60° \) (using the identity \( \sin(180° - \theta) = \sin \theta \))

\( = \frac{\sqrt{3}}{2} \) (since \( \sin 60° = \frac{\sqrt{3}}{2} \))

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: The triple angle formula for sine simplifies \( 3 \sin 40° - \sin^3 40° \) into \( \sin 120° \), which is the same as \( \sin 60° \) due to the supplementary angle property.

Exam Tip: When you see \( 3 \sin x - \sin^3 x \), immediately think of the triple angle formula. Always reduce the resulting angle to a standard value using supplementary or complementary angle identities.

Question 22. Prove that \( \sin^2 24° - \sin^2 6° = \frac{(\sqrt{5} - 1)}{8} \)
Answer: Taking the left side, we begin with \( \sin^2 24° - \sin^2 6° \). Using the difference of squares identity \( \sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B) \):

\( = \sin(24° + 6°) \sin(24° - 6°) \)

\( = \sin 30° \sin 18° \)

Now we must find the value of \( \sin 18° \). Let \( x = 18° \), so \( 5x = 90° \).

We can write \( 2x + 3x = 90° \), which gives \( 2x = 90° - 3x \).

Taking sine of both sides:

\( \sin 2x = \sin(90° - 3x) \)

\( \sin 2x = \cos 3x \) (using the cofunction identity \( \sin(90° - \theta) = \cos \theta \))

We know that \( \sin 2x = 2\sin x \cos x \), \( \cos 3x = 4 \cos^3 x - 3 \cos x \), and \( \cos^2 x + \sin^2 x = 1 \) (so \( \cos^2 x = 1 - \sin^2 x \)).

Substituting \( \sin x = \sin 18° \) and using the identity \( 2\sin x \cos x = 4(1 - \sin^2 x)^{3/2} - 3(1 - \sin^2 x)^{1/2} \) leads to a quartic equation in \( \sin x \). Solving this (or using known trigonometric results):

\( \sin 18° = \frac{\sqrt{5} - 1}{4} \)

Putting this value into our equation:

\( = \sin 30° \sin 18° \)

\( = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \)

\( = \frac{\sqrt{5} - 1}{8} \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: We break the difference of squares into a product of sines at related angles. The key step is determining \( \sin 18° \), which equals \( \frac{\sqrt{5} - 1}{4} \) - a standard exact value worth memorizing.

Exam Tip: The exact values \( \sin 18° = \frac{\sqrt{5} - 1}{4} \) and \( \cos 18° = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \) appear frequently in competition problems. Learn to recognize when an angle can be expressed as a multiple of 18° and use these values accordingly.

Question 23. Prove that \( \sin^2 72° - \cos^2 30° = \frac{(\sqrt{5} - 1)}{8} \)
Answer: Taking the left side, we begin with \( \sin^2 72° - \cos^2 30° \). Since \( 72° = 90° - 18° \), we have \( \sin 72° = \sin(90° - 18°) = \cos 18° \). Thus:

\( = \cos^2 18° - \cos^2 30° \)

Using the identity \( \cos^2 x = 1 - \sin^2 x \):

\( \therefore \cos 18° = \sqrt{1 - \sin^2 18°} \)

\( \Rightarrow \cos 18° = \sqrt{1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2} \)

\( \Rightarrow \cos 18° = \sqrt{\frac{16 - (5 + 1 - 2\sqrt{5})}{16}} \)

\( \Rightarrow \cos 18° = \sqrt{\frac{16 - 6 + 2\sqrt{5}}{16}} \)

\( \Rightarrow \cos 18° = \frac{1}{4}\sqrt{10 + 2\sqrt{5}} \)

Putting the value into the equation, we get:

\( = \cos^2 18° - \cos^2 30° \)

\( = \left(\frac{1}{4}\sqrt{10 + 2\sqrt{5}}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \) (since \( \cos 30° = \frac{\sqrt{3}}{2} \))

\( = \frac{1}{16}(10 + 2\sqrt{5}) - \frac{3}{4} \)

\( = \frac{10 + 2\sqrt{5} - 12}{16} \)

\( = \frac{2\sqrt{5} - 2}{16} \)

\( = \frac{2(\sqrt{5} - 1)}{16} \)

\( = \frac{\sqrt{5} - 1}{8} \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: We first use the cofunction identity to convert \( \sin 72° \) into \( \cos 18° \). Then we apply the Pythagorean identity with the known value of \( \sin 18° \) to find \( \cos 18° \), and finally compute the difference of the squared cosines.

Exam Tip: Problems involving 72° and 18° often require recognizing that these angles are complementary supplements. Use \( \sin 72° = \cos 18° \) and the known exact values for these special angles to reduce complex expressions.

Question 24. Prove that \( \tan 6° \tan 42° \tan 66° \tan 78° = 1 \)
Answer: Taking the left side, we have \( \tan 6° \tan 42° \tan 66° \tan 78° \). Multiply and divide by \( \tan 54° \tan 18° \):

\( = \frac{\tan 6° \tan 42° \tan 66° \tan 78°}{\tan 54° \tan 18°} \times \tan 54° \tan 18° \)

\( = \frac{(\tan 6° \tan 54° \tan 66°)(\tan 18° \tan 42° \tan 72°)}{\tan 54° \tan 18°} \)

Using the identity \( \tan x \tan(60° - x) \tan(60° + x) = \tan 3x \):

For \( x = 6° \): \( \tan 6° \tan(60° - 6°) \tan(60° + 6°) = \tan 6° \tan 54° \tan 66° = \tan 3(6°) = \tan 18° \)

For \( x = 18° \): \( \tan 18° \tan(60° - 18°) \tan(60° + 18°) = \tan 18° \tan 42° \tan 78° = \tan 3(18°) = \tan 54° \)

So the equation becomes:

\( = \frac{[\tan 3(6°)][\tan 3(18°)]}{\tan 54° \tan 18°} \)

\( = \frac{\tan 18° \tan 54°}{\tan 54° \tan 18°} \)

\( = 1 \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: The key is recognizing that angles like 6°, 42°, 66°, and 78° are strategically positioned relative to 60°. The triple angle formula for tangent lets us group them into expressions that cancel out perfectly.

Exam Tip: When you see four tangent terms multiplied together with angles related to multiples of 60°, look for the identity \( \tan x \tan(60° - x) \tan(60° + x) = \tan 3x \). Pairing angles around 60° is the pathway to simplification.

Question 25. If \( \tan \theta = \frac{a}{b} \), prove that \( a \sin 2\theta + b \cos 2\theta = b \)
Answer: Given: \( \tan \theta = \frac{a}{b} \)

To prove: \( a \sin 2\theta + b \cos 2\theta = b \)

From the definition of tangent, we have:

\( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{a}{b} \)

By the Pythagorean theorem:

(Perpendicular)² + (Base)² = (Hypotenuse)²

\( \Rightarrow (a)^2 + (b)^2 = (H)^2 \)

\( \Rightarrow a^2 + b^2 = (H)^2 \)

\( \Rightarrow H = \sqrt{a^2 + b^2} \)

Therefore:

\( \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{\sqrt{a^2 + b^2}} \)

\( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{b}{\sqrt{a^2 + b^2}} \)

Taking the left side of what we need to prove:

\( = a \sin 2\theta + b \cos 2\theta \)

We know that:

\( \sin 2x = 2 \sin x \cos x \)

\( \cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x \)

Using the form \( \cos 2x = \cos^2 x - \sin^2 x \):

\( = a(2 \sin \theta \cos \theta) + b(\cos^2 \theta - \sin^2 \theta) \)

\( = 2a \sin \theta \cos \theta + b \cos^2 \theta - b \sin^2 \theta \)

Substituting the values of \( \sin \theta \) and \( \cos \theta \):

\( = 2a \left(\frac{a}{\sqrt{a^2 + b^2}}\right) \left(\frac{b}{\sqrt{a^2 + b^2}}\right) + b \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2 - b \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 \)

\( = \frac{2ab^2}{a^2 + b^2} + \frac{b^3}{a^2 + b^2} - \frac{ba^2}{a^2 + b^2} \)

\( = \frac{2a^2 b + b^3 - ba^2}{a^2 + b^2} \)

\( = \frac{a^2 b + b^3}{a^2 + b^2} \)

\( = \frac{b(a^2 + b^2)}{a^2 + b^2} \)

\( = b \)

\( = \) RHS

\( \therefore \) LHS \( = \) RHS

Hence Proved
In simple words: Starting from \( \tan \theta = \frac{a}{b} \), we construct a right triangle and use the Pythagorean theorem to find the exact values of \( \sin \theta \) and \( \cos \theta \). Then we substitute these into the double angle formulas and simplify to arrive at the required result.

Exam Tip: Whenever a problem gives a trigonometric ratio in terms of letters (like \( \tan \theta = \frac{a}{b} \)), always draw a right triangle and use the Pythagorean theorem to express all six trig functions. This systematic approach eliminates errors in substitution and simplification.

Exercise 15(E)

Question 1. If sin x = \( \frac{\sqrt{5}}{3} \) and \( \frac{\pi}{2} < x < \pi \), find the values of (i) sin \( \frac{x}{2} \) (ii) cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)
Answer: We are told that sin x = \( \frac{\sqrt{5}}{3} \) and \( \frac{\pi}{2} < x < \pi \), which means x is located in Quadrant II.

To find the values we need: (i) sin \( \frac{x}{2} \) (ii) cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)

First, we determine cos x using the Pythagorean identity:

\[ \text{cos } x = \pm\sqrt{1 - \sin^2 x} \]

\[ \cos x = \pm\sqrt{1 - \left(\frac{\sqrt{5}}{3}\right)^2} = \pm\sqrt{1 - \frac{5}{9}} = \pm\sqrt{\frac{4}{9}} = \pm\frac{2}{3} \]

Since cosine is negative in Quadrant II, we have \( \cos x = -\frac{2}{3} \)

(i) sin \( \frac{x}{2} \)

Using the half-angle formula:

\[ \sin \frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \]

\[ \text{Now, } \sin \frac{x}{2} = \pm\sqrt{\frac{1-\left(-\frac{2}{3}\right)}{2}} = \pm\sqrt{\frac{1+\frac{2}{3}}{2}} = \pm\sqrt{\frac{5}{6}} \]

Since sine is positive in Quadrant II, we get \( \sin \frac{x}{2} = \sqrt{\frac{5}{6}} \)

(ii) cos \( \frac{x}{2} \)

Using the half-angle formula:

\[ \cos \frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}} \]

\[ \text{Now, } \cos \frac{x}{2} = \pm\sqrt{\frac{1+\left(-\frac{2}{3}\right)}{2}} = \pm\sqrt{\frac{1-\frac{2}{3}}{2}} = \pm\sqrt{\frac{1}{6}} \]

Since cosine is negative in Quadrant II, we get \( \cos \frac{x}{2} = -\frac{1}{\sqrt{6}} \)

(iii) tan \( \frac{x}{2} \)

Using the half-angle formula:

\[ \tan x = \frac{\sin x}{\cos x} \]

\[ \text{Hence, } \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{\frac{5}{6}}}{-\frac{1}{\sqrt{6}}} = \frac{\sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{-1} = -\sqrt{5} \]

Since tangent is negative in Quadrant II, this result is consistent.
In simple words: Start by finding cos x using the given sin x and the fact that x is in Quadrant II. Then apply the half-angle formulas for sine, cosine, and tangent. Pay attention to the signs - they depend on which quadrant \( \frac{x}{2} \) falls into.

Exam Tip: Always determine the sign of trigonometric values based on the quadrant. Half-angle formulas give two possible values; choose the correct sign by checking which quadrant the half-angle lies in.

Question 2. If cos x = \( -\frac{3}{5} \) and \( \frac{\pi}{2} < x < \pi \), find the values of (i) sin \( \frac{x}{2} \) (ii) cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)
Answer: We are given that \( \cos x = -\frac{3}{5} \) and \( \frac{\pi}{2} < x < \pi \), meaning x lies in Quadrant II.

To Find: (i) sin \( \frac{x}{2} \) (ii) cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)

(i) sin \( \frac{x}{2} \)

Formula used:

\[ \sin \frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \]

\[ \text{Now, } \sin \frac{x}{2} = \pm\sqrt{\frac{1-\left(-\frac{3}{5}\right)}{2}} = \pm\sqrt{\frac{1+\frac{3}{5}}{2}} = \pm\sqrt{\frac{8}{10}} = \pm\sqrt{\frac{4}{5}} = \pm\frac{2}{\sqrt{5}} \]

Since sine is positive in Quadrant II, we have \( \sin \frac{x}{2} = \frac{2}{\sqrt{5}} \)

(ii) cos \( \frac{x}{2} \)

Formula used:

\[ \cos \frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}} \]

\[ \text{Now, } \cos \frac{x}{2} = \pm\sqrt{\frac{1+\left(-\frac{3}{5}\right)}{2}} = \pm\sqrt{\frac{1-\frac{3}{5}}{2}} = \pm\sqrt{\frac{2}{10}} = \pm\sqrt{\frac{1}{5}} \]

Since cosine is negative in Quadrant II, we get \( \cos \frac{x}{2} = -\frac{1}{\sqrt{5}} \)

(iii) tan \( \frac{x}{2} \)

Formula used:

\[ \tan x = \frac{\sin x}{\cos x} \]

\[ \text{Hence, } \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{2}{\sqrt{5}}}{-\frac{1}{\sqrt{5}}} = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{-1} = -2 \]

Since tangent is negative in Quadrant II, this is consistent with our result.
In simple words: Apply the half-angle formulas for each trigonometric ratio. The sign of each result depends on which quadrant \( \frac{x}{2} \) falls in. Here, tan \( \frac{x}{2} \) comes out negative because tangent is negative when x is in the second half of the interval.

Exam Tip: Always verify that your computed values satisfy the original constraint about which quadrant the angle lies in. Double-check sign choices using the quadrant rule.

Question 3. If Sin x = \( -\frac{1}{2} \) and x lies in Quadrant IV, find the values of (i) Sin \( \frac{x}{2} \) (ii) Cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)
Answer: We are given that \( \sin x = -\frac{1}{2} \) and x is located in Quadrant IV.

To Find: (i) sin \( \frac{x}{2} \) (ii) cos \( \frac{x}{2} \) (iii) tan \( \frac{x}{2} \)

First, we compute cos x using the fundamental trigonometric identity:

\[ \cos x = \pm\sqrt{1 - \sin^2 x} \]

\[ \cos x = \pm\sqrt{1 - \left(-\frac{1}{2}\right)^2} = \pm\sqrt{1 - \frac{1}{4}} = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2} \]

Since cosine is positive in Quadrant IV, we have \( \cos x = \frac{\sqrt{3}}{2} \)

(i) sin \( \frac{x}{2} \)

Formula used:

\[ \sin \frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \]

\[ \text{Now, } \sin \frac{x}{2} = \pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} = \pm\sqrt{\frac{2-\sqrt{3}}{4}} = \pm\frac{\sqrt{2-\sqrt{3}}}{2} \]

Since sine is negative in Quadrant IV, we have \( \sin \frac{x}{2} = -\frac{\sqrt{2-\sqrt{3}}}{2} \)

(ii) cos \( \frac{x}{2} \)

Formula used:

\[ \cos \frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}} \]

\[ \text{Now, } \cos \frac{x}{2} = \pm\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} = \pm\sqrt{\frac{2+\sqrt{3}}{4}} = \pm\frac{\sqrt{2+\sqrt{3}}}{2} \]

Since cosine is positive in Quadrant IV, we get \( \cos \frac{x}{2} = \frac{\sqrt{2+\sqrt{3}}}{2} \)

(iii) tan \( \frac{x}{2} \)

Formula used:

\[ \tan x = \frac{\sin x}{\cos x} \]
In simple words: First determine cos x from the given sin x and the quadrant information. Then use the half-angle formulas, choosing the correct sign for each function based on where \( \frac{x}{2} \) lies.

Exam Tip: In Quadrant IV, sine and tangent are negative while cosine is positive. Apply these sign rules carefully when selecting the correct root from the half-angle formulas.

Question 4. If Cos \( \frac{x}{2} = \frac{12}{13} \) and x lies in Quadrant I, find the values of (i) sin x (ii) cos x (iii) cot x
Answer: We are given that \( \cos \frac{x}{2} = \frac{12}{13} \) and x is located in Quadrant I, meaning all trigonometric values are positive.

To Find: (i) sin x (ii) cos x (iii) cot x

(i) sin x

Formula used:

\[ \sin x = \sqrt{1 - \cos^2 x} \]

We know that \( \cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}} \) (since cosine is positive in Quadrant I)

\[ \Rightarrow 2\cos^2 \frac{x}{2} - 1 = \cos x \]

\[ \Rightarrow 2 \times \left(\frac{12}{13}\right)^2 - 1 = \cos x \]

\[ \Rightarrow 2 \times \frac{144}{169} - 1 = \cos x \]

\[ \Rightarrow \cos x = \frac{119}{169} \]

Since \( \sin x = \sqrt{1 - \cos^2 x} \):

\[ \Rightarrow \sin x = \sqrt{1 - \left(\frac{119}{169}\right)^2} = \frac{120}{169} \]

Hence, we have Sin x = \( \frac{120}{169} \)

(ii) cos x

Formula used:

\[ \text{We know that, } \cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}} \text{ (since cos x is positive in I quadrant)} \]

\[ \Rightarrow 2\cos^2 \frac{x}{2} - 1 = \cos x \]

\[ \Rightarrow 2 \times \left(\frac{12}{13}\right)^2 - 1 = \cos x \]

\[ \Rightarrow \cos x = \frac{119}{169} \]

(iii) cot x

Formula used:

\[ \cot x = \frac{\cos x}{\sin x} \]

\[ \cot x = \frac{\frac{119}{169}}{\frac{120}{169}} = \frac{119}{169} \times \frac{169}{120} = \frac{119}{120} \]

Hence, we have cot x = \( \frac{119}{120} \)
In simple words: From the half-angle value \( \cos \frac{x}{2} \), use the double-angle formula to find cos x. Then apply the Pythagorean identity to get sin x. Finally, divide cos x by sin x to obtain cot x.

Exam Tip: Remember the key formula \( 2\cos^2 \frac{x}{2} - 1 = \cos x \) - this bridges the half-angle and full angle. Always verify that your results are positive when the angle lies in Quadrant I.

Question 5. If sin x = \( \frac{3}{5} \) and \( 0 < x < \frac{\pi}{2} \), find the value of tan \( \frac{x}{2} \)
Answer: We are given that \( \sin x = \frac{3}{5} \) and \( 0 < x < \frac{\pi}{2} \), meaning x is located in Quadrant I and all trigonometric values are positive.

To Find: \( \tan \frac{x}{2} \)

Formula used:

\[ \tan \frac{x}{2} = \frac{\sin x}{1+\cos x} \]

We first find cos x. Since \( \cos x = \sqrt{1 - \sin^2 x} \) (cos is positive in Quadrant I):

\[ \cos x = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

Since \( \tan \frac{x}{2} = \frac{\sin x}{1+\cos x} = \frac{\frac{3}{5}}{1+\frac{4}{5}} = \frac{\frac{3}{5}}{\frac{9}{5}} = \frac{3}{5} \times \frac{5}{9} = \frac{1}{3} \)

Hence, \( \tan \frac{x}{2} = \frac{1}{3} \)
In simple words: Find cos x from the given sin x using the Pythagorean identity. Then substitute both sin x and cos x into the half-angle formula for tangent. The calculation simplifies straightforwardly to give the result.

Exam Tip: The formula \( \tan \frac{x}{2} = \frac{\sin x}{1+\cos x} \) is one of the most practical half-angle identities - it avoids the ± ambiguity and works directly from sin and cos of the full angle.

Question 6. Prove that cot \( \frac{x}{2} \) - tan \( \frac{x}{2} \) = 2 cot x
Answer: To Prove: \( \cot \frac{x}{2} - \tan \frac{x}{2} = 2 \cot x \)

Proof: Consider L.H.S.,

\[ \cot \frac{x}{2} - \tan \frac{x}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} - \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \]

\[ = \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\sin\frac{x}{2} \cos\frac{x}{2}} \]

\[ = \frac{\cos x}{\sin\frac{x}{2} \cos\frac{x}{2}} \text{ (∵ } \cos^2 x - \sin^2 x = \cos 2x\text{)} \]

\[ \Rightarrow \left(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos x\right) \]

Here multiply and divide L.H.S by 2

\[ = \frac{2 \cos x}{2\sin\frac{x}{2} \cos\frac{x}{2}} \]

\[ = \frac{2\cos x}{\sin x} \text{ (∵2sinxcosx = sin2x)} \]

\[ \Rightarrow \left(2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x\right) \]

\[ \cot - \tan\frac{x}{2} = 2\cot x = \text{R.H.S} \]

\[ \therefore \text{L.H.S} = \text{R.H.S, Hence proved} \]
In simple words: Begin by writing cot and tan in terms of sine and cosine. Combine the two fractions over a common denominator. Use the double-angle identities to simplify the numerator and denominator. Multiply numerator and denominator by 2 to introduce the full angle identities, which then give the right-hand side.

Exam Tip: This proof relies heavily on recognizing the double-angle formulas \( \cos 2x = \cos^2 x - \sin^2 x \) and \( \sin 2x = 2\sin x \cos x \). Always keep these identities at hand when proving half-angle identities.

Question 7. Prove that tan\(\left(\frac{\pi}{4} + \frac{x}{2}\right)\) = tan x + sec x
Answer: To Prove: \( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = \tan x + \sec x \)

Proof: Consider L.H.S.,

\[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}} \text{ (∵ this is of the form } \tan(x + y) = \frac{\tan x+\tan y}{1-\tan x\tan y}\text{)} \]

\[ = \frac{1+\tan\frac{x}{2}}{1-\tan \frac{x}{2}} \]

\[ = \frac{1+\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{1-\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} \]

\[ = \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}} \]

Multiply and divide L.H.S by \( \cos\frac{x}{2} + \sin \frac{x}{2} \):

\[ = \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}} \times \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}} \]

\[ = \frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}} \]

\[ = \frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\cos\frac{x}{2}\sin\frac{x}{2}}{\cos x} \text{ (∵ } \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos x\text{)} \]

\[ = \frac{1+2\cos\frac{x}{2}\sin\frac{x}{2}}{\cos x} \]

\[ = \frac{1+\sin x}{\cos x} \]
In simple words: Apply the tangent addition formula for \( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \). Since \( \tan\frac{\pi}{4} = 1 \), this simplifies to a fraction involving \( \tan\frac{x}{2} \). Substitute the definition of tangent in terms of sine and cosine, then rationalize by multiplying by the conjugate. Use double-angle identities to express the result in terms of the full angle x, which yields tan x + sec x.

Exam Tip: Always apply the tangent addition formula first. The conjugate multiplication step is key - it transforms the half-angle expressions into full-angle identities. Remember that \( \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \cos x \) and \( 2\sin\frac{x}{2}\cos\frac{x}{2} = \sin x \).

Question 8. Prove that \( \sqrt{\frac{1 + \sin x}{1 - \sin x}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \)
Answer: Start by taking the left-hand side: \( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \). Multiply both the numerator and denominator by \( \sqrt{1 + \sin x} \):

\[ = \frac{\sqrt{1+\sin x} \times \sqrt{1+\sin x}}{\sqrt{1-\sin x} \times \sqrt{1+\sin x}} = \frac{1+\sin x}{\sqrt{1-\sin^2 x}} \]

Since \( 1 - \sin^2 x = \cos^2 x \), we have:

\[ = \frac{1+\sin x}{\cos x} = \frac{1}{\cos x} + \frac{\sin x}{\cos x} \]

Using the identities \( 1 + 2\cos\frac{x}{2}\sin\frac{x}{2} = \sin x \) and \( \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \cos x \), rewrite the numerator and denominator:

\[ = \frac{\cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\cos\frac{x}{2}\sin\frac{x}{2}}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}} = \frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}} \]

Factor the denominator using the difference of squares formula:

\[ = \frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)} = \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} \]

Now divide both numerator and denominator by \( \cos\frac{x}{2} \):

\[ = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} \]

Since \( \tan\frac{\pi}{4} = 1 \), apply the tangent addition formula:

\[ = \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \]

Therefore, the left-hand side equals the right-hand side, completing the proof.
In simple words: Break down the square root by multiplying top and bottom, then use half-angle identities to rewrite everything in terms of \(\frac{x}{2}\). This gives you the tangent of a sum of two angles, which matches the right side.

Exam Tip: Always start by working on the more complex side (usually the left) and systematically use Pythagorean identities and half-angle formulas. Watch for opportunities to factor and cancel terms to reach the simpler form.

Question 9. Prove that \( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = 2\sec x \)
Answer: Start with the left-hand side: \( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \). Using the tangent addition formula, \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \) and \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):

\[ = \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} + \frac{\tan\frac{\pi}{4} - \tan\frac{x}{2}}{1 + \tan\frac{\pi}{4}\tan\frac{x}{2}} \]

Since \( \tan\frac{\pi}{4} = 1 \), this becomes:

\[ = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} + \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}} \]

Convert \( \tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \):

\[ = \frac{1 + \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{1 - \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} + \frac{1 - \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{1 + \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} \]

\[ = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} + \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}} \]

Combine the two fractions by finding a common denominator:

\[ = \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)} \]

Expand the numerator using the identity \( (a+b)^2 + (a-b)^2 = 2(a^2 + b^2) \):

\[ = \frac{2\left(\cos^2\frac{x}{2} + \sin^2\frac{x}{2}\right)}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}} \]

Since \( \cos^2\frac{x}{2} + \sin^2\frac{x}{2} = 1 \) and \( \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \cos x \):

\[ = \frac{2}{\cos x} = 2\sec x = \text{R.H.S} \]

Thus the left-hand side equals the right-hand side, proving the identity.
In simple words: Apply the tangent addition formula to both terms, notice that they have a symmetric structure because of the plus and minus signs, and then add them by finding a common denominator. The numerator simplifies using a useful algebraic identity, and the denominator reduces to the cosine double-angle formula.

Exam Tip: Watch for symmetry in expressions - when you see \(\tan(A+B) + \tan(A-B)\), look for patterns that allow cancellation after combining fractions. Double-angle identities are key here.

Question 10. Prove that \( \frac{\sin x}{1 + \cos x} = \tan\frac{x}{2} \)
Answer: Begin with the left-hand side: \( \frac{\sin x}{1 + \cos x} \). Using the double-angle identities \( \sin x = 2\cos\frac{x}{2}\sin\frac{x}{2} \) and \( \cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} \):

\[ = \frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{1 + \cos^2\frac{x}{2} - \sin^2\frac{x}{2}} \]

Since \( \cos^2\frac{x}{2} + \sin^2\frac{x}{2} = 1 \), the denominator becomes:

\[ 1 + \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} + \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = 2\cos^2\frac{x}{2} \]

Therefore:

\[ = \frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{2\cos^2\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} = \text{R.H.S} \]

Thus, the left-hand side equals the right-hand side, completing the proof.
In simple words: Replace \(\sin x\) and \(\cos x\) with their half-angle versions, simplify the denominator using the Pythagorean identity, and the fraction reduces directly to the tangent of half the angle.

Exam Tip: Half-angle identities are the key to this proof - always check whether converting to half-angles will simplify the expression. The denominator simplification is the crucial step that makes everything cancel cleanly.