Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 25 Complementary Angles

Exercise 25(A)

Question 1(i). Evaluate: \( \frac{\cos 22^\circ}{\sin 68^\circ} \)
Answer:
By using the formula \( \cos(90^\circ - \theta) = \sin \theta \), we get:
\( \frac{\cos 22^\circ}{\sin 68^\circ} = \frac{\cos(90^\circ - 68^\circ)}{\sin 68^\circ} \)
\( = \frac{\sin 68^\circ}{\sin 68^\circ} \)
\( = 1 \)
In simple words: Change the top cosine term to sine using complementary angles. Since the top and bottom are now the same, they divide to give 1.

Exam Tip: For complementary angles, if the sum of two angles is \( 90^\circ \), convert one trigonometric ratio to match the other so they cancel out.

Question 1(ii). Evaluate: \( \frac{\tan 47^\circ}{\cot 43^\circ} \)
Answer:
Applying the complementary angle identity \( \tan(90^\circ - \theta) = \cot \theta \):
\( \frac{\tan 47^\circ}{\cot 43^\circ} = \frac{\tan(90^\circ - 43^\circ)}{\cot 43^\circ} \)
\( = \frac{\cot 43^\circ}{\cot 43^\circ} \)
\( = 1 \)
In simple words: Turn the tangent term on top into a cotangent term. When the top and bottom match, the fraction equals 1.

Exam Tip: Remember that \( \tan(90^\circ - \theta) = \cot \theta \). Converting the numerator makes it easy to simplify the expression.

Question 1(iii). Evaluate: \( \frac{\sec 75^\circ}{\operatorname{cosec} 15^\circ} \)
Answer:
Using the trigonometric relation \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \):
\( \frac{\sec 75^\circ}{\operatorname{cosec} 15^\circ} = \frac{\sec(90^\circ - 15^\circ)}{\operatorname{cosec} 15^\circ} \)
\( = \frac{\operatorname{cosec} 15^\circ}{\operatorname{cosec} 15^\circ} \)
\( = 1 \)
In simple words: Rewrite secant as cosecant by subtracting the angle from 90 degrees. This leaves equal values on top and bottom, which simplify to 1.

Exam Tip: Check if the two angles add up to 90 degrees. Here, \( 75^\circ + 15^\circ = 90^\circ \), which is a clear sign to use complementary relations.

Question 1(iv). Evaluate: \( \frac{\cos 55^\circ}{\sin 35^\circ} + \frac{\cot 35^\circ}{\tan 55^\circ} \)
Answer:
We can simplify each fraction individually using complementary angle formulas:
\( \frac{\cos 55^\circ}{\sin 35^\circ} + \frac{\cot 35^\circ}{\tan 55^\circ} = \frac{\cos(90^\circ - 35^\circ)}{\sin 35^\circ} + \frac{\cot(90^\circ - 55^\circ)}{\tan 55^\circ} \)
Using \( \cos(90^\circ - \theta) = \sin \theta \) and \( \cot(90^\circ - \theta) = \tan \theta \):
\( = \frac{\sin 35^\circ}{\sin 35^\circ} + \frac{\tan 55^\circ}{\tan 55^\circ} \)
\( = 1 + 1 \)
\( = 2 \)
In simple words: Change cosine to sine in the first part, and cotangent to tangent in the second part. Both fractions become 1, and adding them gives 2.

Exam Tip: When you have multiple fractions, work on each one step-by-step. Convert only one term in each fraction to find the simple values.

Question 1(v). Evaluate: \( \sin^2 40^\circ - \cos^2 50^\circ \)
Answer:
We know that \( \sin(90^\circ - \theta) = \cos \theta \). Let us convert the first term:
\( \sin^2 40^\circ - \cos^2 50^\circ = \sin^2(90^\circ - 50^\circ) - \cos^2 50^\circ \)
\( = \cos^2 50^\circ - \cos^2 50^\circ \)
\( = 0 \)
In simple words: Rewrite the sine squared term using its complementary cosine squared form. Since the two identical values subtract from each other, the result is 0.

Exam Tip: Keep track of squares when applying complementary formulas; the squared symbol remains on the converted trigonometric ratio.

Question 1(vi). Evaluate: \( \sec^2 18^\circ - \operatorname{cosec}^2 72^\circ \)
Answer:
Convert the secant term using \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \):
\( \sec^2 18^\circ - \operatorname{cosec}^2 72^\circ = [\sec(90^\circ - 72^\circ)]^2 - \operatorname{cosec}^2 72^\circ \)
\( = \operatorname{cosec}^2 72^\circ - \operatorname{cosec}^2 72^\circ \)
\( = 0 \)
In simple words: Convert the first term into its cosecant equivalent. Subtracting the two identical terms leaves 0.

Exam Tip: Remember that \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \). Write the intermediate step carefully to avoid losing the square exponent.

Question 1(vii). Evaluate: \( \sin 15^\circ \cos 15^\circ - \cos 75^\circ \sin 75^\circ \)
Answer:
We can express the terms using complementary relations. Let us convert \( \sin 15^\circ \) and \( \sin 75^\circ \):
\( \sin 15^\circ \cos 15^\circ - \cos 75^\circ \sin 75^\circ \)
\( = \sin(90^\circ - 75^\circ)\cos 15^\circ - \cos 75^\circ \sin(90^\circ - 15^\circ) \)
Using \( \sin(90^\circ - \theta) = \cos \theta \):
\( = \cos 75^\circ \cos 15^\circ - \cos 75^\circ \cos 15^\circ \)
\( = 0 \)
In simple words: Change the sine terms into cosine terms using complementary angles. The two sides of the subtraction sign then match exactly, giving a final answer of 0.

Exam Tip: When you have products of terms, make sure to apply the identities consistently so that matching terms are formed on both sides of the subtraction sign.

Question 1(viii). Evaluate: \( \sin 42^\circ \sin 48^\circ - \cos 42^\circ \cos 48^\circ \)
Answer:
Convert the angles in the first part of each term to their complements:
\( \sin 42^\circ \sin 48^\circ - \cos 42^\circ \cos 48^\circ \)
\( = \sin(90^\circ - 48^\circ)\sin 48^\circ - \cos(90^\circ - 48^\circ)\cos 48^\circ \)
Using complementary angle formulas, we get:
\( = \cos 48^\circ \sin 48^\circ - \sin 48^\circ \cos 48^\circ \)
Rearranging the second term:
\( = \cos 48^\circ \sin 48^\circ - \cos 48^\circ \sin 48^\circ \)
\( = 0 \)
In simple words: Rewrite the first angle in both sections to 48 degrees using complementary formulas. This makes both sides of the minus sign identical, so they cancel out to 0.

Exam Tip: Rearrange terms alphabetically or in a standard order after conversion to clearly see when they cancel out.

Question 2(i). Evaluate: \( \sin(90^\circ - A)\sin A - \cos(90^\circ - A)\cos A \)
Answer:
Apply the identities \( \sin(90^\circ - A) = \cos A \) and \( \cos(90^\circ - A) = \sin A \):
\( \sin(90^\circ - A)\sin A - \cos(90^\circ - A)\cos A \)
\( = \cos A \sin A - \sin A \cos A \)
Since multiplication is commutative, \( \sin A \cos A = \cos A \sin A \):
\( = 0 \)
In simple words: Change the complementary parts into plain cosine and sine terms. Because the two halves are exactly the same, subtracting them leaves 0.

Exam Tip: Remember that \( \sin A \cos A \) and \( \cos A \sin A \) are equal because order does not matter in multiplication.

Question 2(ii). Evaluate: \( \sin^2 35^\circ - \cos^2 55^\circ \)
Answer:
Let us convert the second term using \( \cos(90^\circ - \theta) = \sin \theta \):
\( \sin^2 35^\circ - \cos^2 55^\circ = \sin^2 35^\circ - [\cos(90^\circ - 35^\circ)]^2 \)
\( = \sin^2 35^\circ - \sin^2 35^\circ \)
\( = 0 \)
In simple words: Swap the second term to its sine squared counterpart. Subtracting the matching terms gives 0.

Exam Tip: You can convert either term to match the other. Converting the second term works just as well as converting the first.

Question 2(iii). Evaluate: \( \frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 \)
Answer:
Use the complementary relationships \( \cot(90^\circ - \theta) = \tan \theta \) and \( \tan(90^\circ - \theta) = \cot \theta \):
\( \frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 = \frac{\cot(90^\circ - 36^\circ)}{\tan 36^\circ} + \frac{\tan(90^\circ - 70^\circ)}{\cot 70^\circ} - 2 \)
\( = \frac{\tan 36^\circ}{\tan 36^\circ} + \frac{\cot 70^\circ}{\cot 70^\circ} - 2 \)
\( = 1 + 1 - 2 \)
\( = 2 - 2 \)
\( = 0 \)
In simple words: Convert the top terms of both fractions using complementary angles. This turns both fractions into 1. Adding them and subtracting 2 leaves 0.

Exam Tip: Be careful with signs and constants like -2 at the end. Perform addition before subtraction following standard order of operations.

Question 2(iv). Evaluate: \( \frac{2\tan 53^\circ}{\cot 37^\circ} - \frac{\cot 80^\circ}{\tan 10^\circ} \)
Answer:
Applying complementary relations to the numerator of each term:
\( \frac{2\tan 53^\circ}{\cot 37^\circ} - \frac{\cot 80^\circ}{\tan 10^\circ} = \frac{2\tan(90^\circ - 37^\circ)}{\cot 37^\circ} - \frac{\cot(90^\circ - 10^\circ)}{\tan 10^\circ} \)
Using \( \tan(90^\circ - \theta) = \cot \theta \) and \( \cot(90^\circ - \theta) = \tan \theta \):
\( = \frac{2\cot 37^\circ}{\cot 37^\circ} - \frac{\tan 10^\circ}{\tan 10^\circ} \)
\( = 2 - 1 \)
\( = 1 \)
In simple words: Rewrite the top parts using complementary angles. The first fraction simplifies to 2, and the second simplifies to 1. Subtracting them gives 1.

Exam Tip: Notice the coefficient 2 in front of the first fraction. Ensure it remains multiplied with the simplified term.

Question 2(v). Evaluate: \( \cos^2 25^\circ - \sin^2 65^\circ - \tan^2 45^\circ \)
Answer:
Use complementary angle formulas and the standard value \( \tan 45^\circ = 1 \):
\( \cos^2 25^\circ - \sin^2 65^\circ - \tan^2 45^\circ = [\cos(90^\circ - 65^\circ)]^2 - \sin^2 65^\circ - (\tan 45^\circ)^2 \)
\( = \sin^2 65^\circ - \sin^2 65^\circ - (1)^2 \)
\( = 0 - 1 \)
\( = -1 \)
In simple words: Convert the first term to sine squared. Since \( \tan 45^\circ \) is 1, the expression simplifies to 0 minus 1, which equals -1.

Exam Tip: Remember standard trigonometric values like \( \tan 45^\circ = 1 \) and combine them with complementary angle rules.

Question 2(vi). Evaluate: \( \left(\frac{\sin 77^\circ}{\cos 13^\circ}\right)^2 + \left(\frac{\cos 77^\circ}{\sin 13^\circ}\right)^2 - 2\cos^2 45^\circ \)
Answer:
Substitute the complementary conversions for the numerators and the standard value \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\( \left(\frac{\sin 77^\circ}{\cos 13^\circ}\right)^2 + \left(\frac{\cos 77^\circ}{\sin 13^\circ}\right)^2 - 2\cos^2 45^\circ \)
\( = \left(\frac{\sin(90^\circ - 13^\circ)}{\cos 13^\circ}\right)^2 + \left(\frac{\cos(90^\circ - 13^\circ)}{\sin 13^\circ}\right)^2 - 2(\cos 45^\circ)^2 \)
\( = \left(\frac{\cos 13^\circ}{\cos 13^\circ}\right)^2 + \left(\frac{\sin 13^\circ}{\sin 13^\circ}\right)^2 - 2\left(\frac{1}{\sqrt{2}}\right)^2 \)
\( = (1)^2 + (1)^2 - 2 \times \frac{1}{2} \)
\( = 1 + 1 - 1 \)
\( = 1 \)
In simple words: Simplify the brackets first by rewriting them with matching angles. This turns them into 1. Substitute the standard value for the last term, then calculate the final sum.

Exam Tip: Be careful when squaring fractions. Since \( \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \), multiplying by 2 yields exactly 1.

Question 3(i). Prove that: \( \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ = 1 \)
Answer:
Let us simplify the Left Hand Side (L.H.S.):
L.H.S. \( = \tan 10^\circ \tan 15^\circ \tan 75^\circ \tan 80^\circ \)
Rewrite the first two terms using complementary angles:
\( = \tan(90^\circ - 80^\circ) \tan(90^\circ - 75^\circ) \tan 75^\circ \tan 80^\circ \)
Using \( \tan(90^\circ - \theta) = \cot \theta \):
\( = \cot 80^\circ \cot 75^\circ \tan 75^\circ \tan 80^\circ \)
Grouping the reciprocal pairs together:
\( = (\cot 80^\circ \tan 80^\circ)(\cot 75^\circ \tan 75^\circ) \)
Since \( \cot \theta \tan \theta = 1 \):
\( = (1)(1) \)
\( = 1 = \text{R.H.S.} \)
Hence proved.
In simple words: Convert the first two tangent terms into cotangents. Grouping them with their matching tangent pairs results in products of 1, which multiply to 1.

Exam Tip: Group reciprocal pairs like \( \tan \theta \) and \( \cot \theta \) together, as their product is always 1.

Question 3(ii). Prove that: \( \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ = 2 \)
Answer:
Starting with the Left Hand Side (L.H.S.):
L.H.S. \( = \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ \)
Express secant and cosecant as reciprocals, and rewrite \( 42^\circ \) using complementary angles:
\( = \sin(90^\circ - 48^\circ) \times \frac{1}{\cos 48^\circ} + \cos(90^\circ - 48^\circ) \times \frac{1}{\sin 48^\circ} \)
Since \( \sin(90^\circ - \theta) = \cos \theta \) and \( \cos(90^\circ - \theta) = \sin \theta \):
\( = \cos 48^\circ \times \frac{1}{\cos 48^\circ} + \sin 48^\circ \times \frac{1}{\sin 48^\circ} \)
\( = 1 + 1 \)
\( = 2 = \text{R.H.S.} \)
Hence proved.
In simple words: Convert secant and cosecant to their reciprocal fraction forms. Change the remaining terms using complementary angles, so they cancel out to 1. Adding them gives 2.

Exam Tip: Convert reciprocal functions like \( \sec \theta = \frac{1}{\cos \theta} \) and \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \) first to simplify products.

Question 4. Express each of the following in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \):
(i) \( \sin 59^\circ + \tan 63^\circ \)
(ii) \( \operatorname{cosec} 68^\circ + \cot 72^\circ \)
(iii) \( \cos 74^\circ + \sec 67^\circ \)
Answer:
(i) Convert using complementary angles:
\( \sin 59^\circ + \tan 63^\circ = \sin(90^\circ - 31^\circ) + \tan(90^\circ - 27^\circ) \)
\( = \cos 31^\circ + \cot 27^\circ \)

(ii) Convert both terms to angles below \( 45^\circ \):
\( \operatorname{cosec} 68^\circ + \cot 72^\circ = \operatorname{cosec}(90^\circ - 22^\circ) + \cot(90^\circ - 18^\circ) \)
\( = \sec 22^\circ + \tan 18^\circ \)

(iii) Rewrite each term using complementary ratios:
\( \cos 74^\circ + \sec 67^\circ = \cos(90^\circ - 16^\circ) + \sec(90^\circ - 23^\circ) \)
\( = \sin 16^\circ + \operatorname{cosec} 23^\circ \)
In simple words: Rewrite each angle as 90 minus a smaller angle. Then use the complementary formulas to switch the trig functions, making the angles smaller than 45 degrees.

Exam Tip: To represent angles between \( 0^\circ \) and \( 45^\circ \), always subtract the given angle from \( 90^\circ \) and use the corresponding complementary ratio.

Question 5. If A, B, and C are the interior angles of a triangle ABC, show that:
(i) \( \sin\left(\frac{A + B}{2}\right) = \cos\left(\frac{C}{2}\right) \)
(ii) \( \tan\left(\frac{B + C}{2}\right) = \cot\left(\frac{A}{2}\right) \)
Answer:
(i) For any triangle \( \Delta ABC \), the sum of the interior angles is \( 180^\circ \):
\( A + B + C = 180^\circ \)
Dividing the equation by 2, we get:
\( \frac{A + B + C}{2} = 90^\circ \)

\( \implies \frac{A + B}{2} = 90^\circ - \frac{C}{2} \)
Taking sine on both sides:
\( \sin\left(\frac{A + B}{2}\right) = \sin\left(90^\circ - \frac{C}{2}\right) \)
Using \( \sin(90^\circ - \theta) = \cos \theta \):
\( \sin\left(\frac{A + B}{2}\right) = \cos\left(\frac{C}{2}\right) \)

(ii) Similarly, for triangle \( \Delta ABC \):
\( B + C = 180^\circ - A \)
Dividing by 2:
\( \frac{B + C}{2} = 90^\circ - \frac{A}{2} \)
Taking tangent on both sides:
\( \tan\left(\frac{B + C}{2}\right) = \tan\left(90^\circ - \frac{A}{2}\right) \)
Since \( \tan(90^\circ - \theta) = \cot \theta \):
\( \tan\left(\frac{B + C}{2}\right) = \cot\left(\frac{A}{2}\right) \)
In simple words: Since all angles in a triangle add up to 180 degrees, half of the sum of two angles is equal to 90 degrees minus half of the third angle. Using complementary angle formulas gives the final results.

Exam Tip: Always start proofs involving interior angles of a triangle by stating the angle sum property: \( A + B + C = 180^\circ \).

Question 6(i). Evaluate: \( 3\frac{\sin 72^\circ}{\cos 18^\circ} - \frac{\sec 32^\circ}{\operatorname{cosec} 58^\circ} \)
Answer:
Use complementary angle conversions for the numerators:
\( 3\frac{\sin 72^\circ}{\cos 18^\circ} - \frac{\sec 32^\circ}{\operatorname{cosec} 58^\circ} = 3\frac{\sin(90^\circ - 18^\circ)}{\cos 18^\circ} - \frac{\sec(90^\circ - 58^\circ)}{\operatorname{cosec} 58^\circ} \)
Since \( \sin(90^\circ - \theta) = \cos \theta \) and \( \sec(90^\circ - \theta) = \operatorname{cosec} \theta \):
\( = 3\frac{\cos 18^\circ}{\cos 18^\circ} - \frac{\operatorname{cosec} 58^\circ}{\operatorname{cosec} 58^\circ} \)
\( = 3(1) - 1 \)
\( = 2 \)
In simple words: Turn the top terms into their complementary forms so they match the denominators. The fractions simplify to 1, leaving us with 3 minus 1, which equals 2.

Exam Tip: Remember to multiply the simplified fraction value by the coefficient outside. Here, \( 3(1) = 3 \).

Question 6(ii). Evaluate: \( 3\cos 80^\circ \operatorname{cosec} 10^\circ + 2\cos 59^\circ \operatorname{cosec} 31^\circ \)
Answer:
Convert the cosine terms using complementary angles:
\( 3\cos 80^\circ \operatorname{cosec} 10^\circ + 2\cos 59^\circ \operatorname{cosec} 31^\circ \)
\( = 3\cos(90^\circ - 10^\circ)\operatorname{cosec} 10^\circ + 2\cos(90^\circ - 31^\circ)\operatorname{cosec} 31^\circ \)
Since \( \cos(90^\circ - \theta) = \sin \theta \):
\( = 3\sin 10^\circ \operatorname{cosec} 10^\circ + 2\sin 31^\circ \operatorname{cosec} 31^\circ \)
Because \( \sin \theta \operatorname{cosec} \theta = 1 \):
\( = 3(1) + 2(1) \)
\( = 5 \)
In simple words: Change the cosine terms to sine. Since sine multiplied by cosecant of the same angle is 1, the expression simplifies to 3 plus 2, which is 5.

Exam Tip: Recognize that \( \sin \theta \) and \( \operatorname{cosec} \theta \) are reciprocals, meaning their product is always 1.

Question 6(iii). Evaluate: \( \frac{\sin 80^\circ}{\cos 10^\circ} + \sin 59^\circ \sec 31^\circ \)
Answer:
Applying the complementary identities \( \sin(90^\circ - \theta) = \cos \theta \) and \( \sec \theta = \frac{1}{\cos \theta} \):
\( \frac{\sin 80^\circ}{\cos 10^\circ} + \sin 59^\circ \sec 31^\circ = \frac{\sin(90^\circ - 10^\circ)}{\cos 10^\circ} + \sin(90^\circ - 31^\circ)\sec 31^\circ \)
\( = \frac{\cos 10^\circ}{\cos 10^\circ} + \cos 31^\circ \times \frac{1}{\cos 31^\circ} \)
\( = 1 + 1 \)
\( = 2 \)
In simple words: Rewrite the sine terms in both parts using complementary angles. This turns each part into 1, giving a sum of 2.

Exam Tip: Remember to convert secant to \( \frac{1}{\cos} \) to clearly show how the terms divide to 1.

Question 6(iv). Evaluate: \( \tan(55^\circ - A) - \cot(35^\circ + A) \)
Answer:
Express \( 55^\circ - A \) as \( 90^\circ - (35^\circ + A) \):
\( \tan(55^\circ - A) - \cot(35^\circ + A) = \tan[90^\circ - (35^\circ + A)] - \cot(35^\circ + A) \)
Using \( \tan(90^\circ - \theta) = \cot \theta \):
\( = \cot(35^\circ + A) - \cot(35^\circ + A) \)
\( = 0 \)
In simple words: Rewrite the angle of the tangent term to match the cotangent angle. Since they are equal and subtracted, the result is 0.

Exam Tip: For expressions with variables like \( A \), group them together with the numerical angles inside the complementary bracket.

Question 6(v). Evaluate: \( \operatorname{cosec}(65^\circ + A) - \sec(25^\circ - A) \)
Answer:
Use the complementary angle identity for cosecant:
\( \operatorname{cosec}(65^\circ + A) - \sec(25^\circ - A) = \operatorname{cosec}[90^\circ - (25^\circ - A)] - \sec(25^\circ - A) \)
Using \( \operatorname{cosec}(90^\circ - \theta) = \sec \theta \):
\( = \sec(25^\circ - A) - \sec(25^\circ - A) \)
\( = 0 \)
In simple words: Rewrite the cosecant term so it becomes a secant term with the same angle. Subtracting them gives 0.

Exam Tip: Confirm that \( (65^\circ + A) + (25^\circ - A) = 90^\circ \). This shows that the two angles are indeed complementary.

Question 6(vi). Evaluate: \( 2\frac{\tan 57^\circ}{\cot 33^\circ} - \frac{\cot 70^\circ}{\tan 20^\circ} - \sqrt{2}\cos 45^\circ \)
Answer:
Use complementary angle formulas and substitute the value \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\( 2\frac{\tan 57^\circ}{\cot 33^\circ} - \frac{\cot 70^\circ}{\tan 20^\circ} - \sqrt{2}\cos 45^\circ \)
\( = 2\frac{\tan(90^\circ - 33^\circ)}{\cot 33^\circ} - \frac{\cot(90^\circ - 20^\circ)}{\tan 20^\circ} - \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \)
Applying \( \tan(90^\circ - \theta) = \cot \theta \) and \( \cot(90^\circ - \theta) = \tan \theta \):
\( = 2\frac{\cot 33^\circ}{\cot 33^\circ} - \frac{\tan 20^\circ}{\tan 20^\circ} - 1 \)
\( = 2(1) - 1 - 1 \)
\( = 0 \)
In simple words: Convert the top terms of both fractions using complementary angles. Put in the value for \( \cos 45^\circ \). This simplifies the problem to 2 minus 1 minus 1, which equals 0.

Exam Tip: Be careful with the multiplication \( \sqrt{2} \times \frac{1}{\sqrt{2}} \). Since they are reciprocals, the product is exactly 1.

Question 6(vii). Evaluate: \( \frac{\cot^2 41^\circ}{\tan^2 49^\circ} - 2\frac{\sin^2 75^\circ}{\cos^2 15^\circ} \)
Answer:
Rewrite the numerators using complementary properties:
\( \frac{\cot^2 41^\circ}{\tan^2 49^\circ} - 2\frac{\sin^2 75^\circ}{\cos^2 15^\circ} = \frac{[\cot(90^\circ - 49^\circ)]^2}{\tan^2 49^\circ} - 2\frac{[\sin(90^\circ - 15^\circ)]^2}{\cos^2 15^\circ} \)
Using \( \cot(90^\circ - \theta) = \tan \theta \) and \( \sin(90^\circ - \theta) = \cos \theta \):
\( = \frac{\tan^2 49^\circ}{\tan^2 49^\circ} - 2\frac{\cos^2 15^\circ}{\cos^2 15^\circ} \)
\( = 1 - 2(1) \)
\( = -1 \)
In simple words: Convert the top part of each fraction to match the bottom part. This turns the fractions into 1, leaving us with 1 minus 2, which equals -1.

Exam Tip: Ensure that when converting terms with squares, the squaring is applied correctly to the complementary function.

Question 6(viii). Evaluate: \( \frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 59^\circ}{\sin 31^\circ} - 8\sin^2 30^\circ \)
Answer:
Apply complementary angle identities and substitute the standard value \( \sin 30^\circ = \frac{1}{2} \):
\( \frac{\cos 70^\circ}{\sin 20^\circ} + \frac{\cos 59^\circ}{\sin 31^\circ} - 8\sin^2 30^\circ \)
\( = \frac{\cos(90^\circ - 20^\circ)}{\sin 20^\circ} + \frac{\cos(90^\circ - 31^\circ)}{\sin 31^\circ} - 8\left(\frac{1}{2}\right)^2 \)
Since \( \cos(90^\circ - \theta) = \sin \theta \):
\( = \frac{\sin 20^\circ}{\sin 20^\circ} + \frac{\sin 31^\circ}{\sin 31^\circ} - 8\left(\frac{1}{4}\right) \)
\( = 1 + 1 - 2 \)
\( = 0 \)
In simple words: Rewrite the cosine terms on top using complementary angles. Use the known value of \( \sin 30^\circ \). This simplifies the problem to 1 plus 1 minus 2, which is 0.

Exam Tip: Always write the value of standard angles like \( 30^\circ, 45^\circ \) separately before squaring to ensure accuracy.

Question 6(ix). Evaluate: \( 14\sin 30^\circ + 6\cos 60^\circ - 5\tan 45^\circ \)
Answer:
Substitute the standard trigonometric values \( \sin 30^\circ = \frac{1}{2} \), \( \cos 60^\circ = \frac{1}{2} \), and \( \tan 45^\circ = 1 \):
\( 14\sin 30^\circ + 6\cos 60^\circ - 5\tan 45^\circ \)
\( = 14\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 5(1) \)
\( = 7 + 3 - 5 \)
\( = 5 \)
In simple words: Put the standard values of the angles into the expression. This gives 7 plus 3 minus 5, which equals 5.

Exam Tip: This question does not require complementary formulas because all angles are standard trigonometric angles.

Question 7. If ABC is a right-angled triangle, right-angled at B, find the value of: \( \frac{\sec A \sin C - \tan A \tan C}{\sin B} \)
Answer:
In a right-angled triangle \( \Delta ABC \) with the right angle at \( B \), we have \( B = 90^\circ \).
The sum of angles in a triangle is \( 180^\circ \), so:
\( A + B + C = 180^\circ \)
\( A + C + 90^\circ = 180^\circ \)

\( \implies A + C = 90^\circ \)

\( \implies A = 90^\circ - C \)
Now, let us substitute \( A = 90^\circ - C \) and \( B = 90^\circ \) into the given expression:
\( \frac{\sec A \sin C - \tan A \tan C}{\sin B} = \frac{\sec(90^\circ - C)\sin C - \tan(90^\circ - C)\tan C}{\sin 90^\circ} \)
Using the complementary angle identities \( \sec(90^\circ - C) = \operatorname{cosec} C \), \( \tan(90^\circ - C) = \cot C \), and \( \sin 90^\circ = 1 \):
\( = \frac{\operatorname{cosec} C \sin C - \cot C \tan C}{1} \)
Since \( \operatorname{cosec} C = \frac{1}{\sin C} \) and \( \cot C = \frac{1}{\tan C} \):
\( = \left(\frac{1}{\sin C} \times \sin C\right) - \left(\frac{1}{\tan C} \times \tan C\right) \)
\( = 1 - 1 \)
\( = 0 \)
In simple words: Since angle B is 90 degrees, the other two angles must add up to 90 degrees. This allows us to convert angle A into 90 minus C. Using reciprocal properties, the terms cancel out to 0.

Exam Tip: Since it is a right triangle, use the relationship \( A + C = 90^\circ \) to substitute \( A \) with its complement in terms of \( C \).

Question 8(i). Find the value of A if \( \sin(90^\circ - 3A)\operatorname{cosec} 42^\circ = 1 \).
Answer:
Given equation:
\( \sin(90^\circ - 3A)\operatorname{cosec} 42^\circ = 1 \)
We can rewrite this by moving cosecant to the right-hand side:

\( \implies \sin(90^\circ - 3A) = \frac{1}{\operatorname{cosec} 42^\circ} \)
Using \( \sin(90^\circ - \theta) = \cos \theta \):

\( \implies \cos 3A = \sin 42^\circ \)
Since \( \sin 42^\circ = \sin(90^\circ - 48^\circ) = \cos 48^\circ \):

\( \implies \cos 3A = \cos 48^\circ \)
Equating the angles:

\( \implies 3A = 48^\circ \)

\( \implies A = 16^\circ \)
In simple words: Use reciprocal and complementary angle relations to match both sides of the equation as cosine functions. Solving for A gives 16 degrees.

Exam Tip: Always try to express both sides of the equation using the same trigonometric ratio (like cosine) to easily equate the angles.

Question 8(ii). Find the value of A if \( \cos(90^\circ - 3A)\sec 77^\circ = 1 \).
Answer:
Given:
\( \cos(90^\circ - 3A)\sec 77^\circ = 1 \)
Dividing both sides by secant:

\( \implies \cos(90^\circ - 3A) = \frac{1}{\sec 77^\circ} \)
Since \( \cos(90^\circ - \theta) = \sin \theta \):

\( \implies \sin 3A = \cos 77^\circ \)
Convert \( \cos 77^\circ \) using complementary angle rules:
\( \cos 77^\circ = \cos(90^\circ - 12^\circ) = \sin 12^\circ \)
Thus:

\( \implies \sin 3A = \sin 12^\circ \)
Comparing the angles:

\( \implies 3A = 12^\circ \)

\( \implies A = 3^\circ \)
In simple words: Move the secant term to the right-hand side. Convert both sides to sine functions, then solve the equation to find that A is 3 degrees.

Exam Tip: Ensure that you perform the divisions carefully and use the exact complementary formulas for secant and cosecant.