RS Aggarwal Solutions for Class 12 Chapter 04 Inverse Trigonometric Functions

Question 1. Find the principal value of:
(i) \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
(ii) \( \sin^{-1}\left(\frac{1}{2}\right) \)
(iii) \( \cos^{-1}\left(\frac{1}{2}\right) \)
(iv) \( \tan^{-1}(1) \)
(v) \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
(vi) \( \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
(vii) \( \csc^{-1}(\sqrt{2}) \)
Answer: When finding the principal value of an inverse trigonometric function, we need to identify which angle produces the given trigonometric result.
(i) Set \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = x \). Then \( \frac{\sqrt{3}}{2} = \sin x \). From the trigonometric table, the angle whose sine equals \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \).

\( \therefore x = \frac{\pi}{3} \)
(ii) Set \( \sin^{-1}\left(\frac{1}{2}\right) = x \). Then \( \frac{1}{2} = \sin x \). The angle whose sine is \( \frac{1}{2} \) is \( \frac{\pi}{6} \).

\( \therefore x = \frac{\pi}{6} \)
(iii) Set \( \cos^{-1}\left(\frac{1}{2}\right) = x \). Then \( \frac{1}{2} = \cos x \). The angle whose cosine equals \( \frac{1}{2} \) is \( \frac{\pi}{3} \).

\( \therefore x = \frac{\pi}{3} \)
(iv) Set \( \tan^{-1}(1) = x \). Then \( 1 = \tan x \). The angle whose tangent is 1 is \( \frac{\pi}{4} \).

\( \therefore x = \frac{\pi}{4} \)
(v) Set \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = x \). Then \( \frac{1}{\sqrt{3}} = \tan x \). The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is \( \frac{\pi}{6} \).

\( \therefore x = \frac{\pi}{6} \)
(vi) Set \( \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = x \). Then \( \frac{2}{\sqrt{3}} = \sec x \). The angle whose secant is \( \frac{2}{\sqrt{3}} \) is \( \frac{\pi}{6} \).

\( \therefore x = \frac{\pi}{6} \)
(vii) Set \( \csc^{-1}(\sqrt{2}) = x \). Then \( \sqrt{2} = \csc x \). The angle whose cosecant equals \( \sqrt{2} \) is \( \frac{\pi}{4} \).

\( \therefore x = \frac{\pi}{4} \)

Exam Tip: Always refer to the standard trigonometric values table to identify which angle matches the given value. The principal value must lie within the specified range for each inverse function.

Question 2. Find the principal value of:
(i) \( \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \)
(ii) \( \cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) \)
(iii) \( \tan^{-1}(-\sqrt{3}) \)
(iv) \( \sec^{-1}(-2) \)
(v) \( \csc^{-1}(-\sqrt{2}) \)
(vi) \( \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) \)
Answer: When the argument is negative, we apply the corresponding inverse function formulas for negative inputs.
(i) Using the formula \( \sin^{-1}(-x) = -\sin^{-1}(x) \):

\( \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) = -\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4} \)

\( \therefore x = -\frac{\pi}{4} \)
(ii) Using the formula \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):

First, set \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \)

\( \cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

\( \therefore x = \frac{5\pi}{6} \)
(iii) Using the formula \( \tan^{-1}(-x) = -\tan^{-1}(x) \):

\( \tan^{-1}(-\sqrt{3}) = -\tan^{-1}(\sqrt{3}) = -\frac{\pi}{3} \)

\( \therefore x = -\frac{\pi}{3} \)
(iv) Using the formula \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \):

First, set \( \sec^{-1}(2) = \frac{\pi}{3} \)

\( \sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

\( \therefore x = \frac{2\pi}{3} \)
(v) Using the formula \( \csc^{-1}(-x) = -\csc^{-1}(x) \):

\( \csc^{-1}(-\sqrt{2}) = -\csc^{-1}(\sqrt{2}) = -\frac{\pi}{4} \)

\( \therefore x = -\frac{\pi}{4} \)
(vi) Using the formula \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \):

First, set \( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} \)

\( \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

\( \therefore x = \frac{2\pi}{3} \)

Exam Tip: Memorize the transformation formulas for negative arguments in inverse trigonometric functions - they are key to solving these quickly and accurately.

Question 3. Evaluate \( \cos\left[\cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) + \frac{\pi}{6}\right] \)
Answer: From the previous question, we know that \( \cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) = \frac{5\pi}{6} \).

Substituting this value:

\( \cos\left[\frac{5\pi}{6} + \frac{\pi}{6}\right] = \cos(\pi) = -1 \)

\( \therefore \text{The answer is } -1 \)

Exam Tip: Always compute the inverse function value first, then apply the outer trigonometric function. Breaking the problem into smaller steps reduces errors.

Question 4. Evaluate \( \sin\left[\frac{\pi}{2} - \left(-\frac{\pi}{3}\right)\right] \)
Answer: Simplify the expression inside the brackets:

\( \sin\left[\frac{\pi}{2} + \frac{\pi}{3}\right] = \sin\left(\frac{5\pi}{6}\right) \)

This angle can be rewritten as:

\( \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)

\( \therefore \text{The answer is } \frac{1}{2} \)

Exam Tip: Use angle addition identities and complementary angle relationships to simplify trigonometric expressions efficiently.

Exercise 4B

Question 1. Find the principal value of each of the following:
\( \sin^{-1}\left(\frac{-1}{2}\right) \)
Answer: Using the formula \( \sin^{-1}(-x) = -\sin^{-1}(x) \):

\( \sin^{-1}\left(\frac{-1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6} \)

Exam Tip: The negative argument formula is the fastest route - apply it directly without hesitation.

Question 2. Find the principal value of each of the following:
\( \cos^{-1}\left(\frac{-1}{2}\right) \)
Answer: Using the formula \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):

\( \cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

Exam Tip: For cosine and other reciprocal functions with negative inputs, subtract the positive argument result from \( \pi \) or use the appropriate supplementary formula.

Question 3. Find the principal value of each of the following:
\( \tan^{-1}(-1) \)
Answer: Using the formula \( \tan^{-1}(-x) = -\tan^{-1}(x) \):

\( \tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4} \)

We recognize that \( \tan\left(\frac{\pi}{4}\right) = 1 \), so \( \tan^{-1}(1) = \frac{\pi}{4} \).

Exam Tip: Recall the standard inverse values for commonly used angles like \( \frac{\pi}{4} \), \( \frac{\pi}{6} \), and \( \frac{\pi}{3} \) to speed up computation.

Question 4. Find the principal value of each of the following:
\( \sec^{-1}(-2) \)
Answer: Using the formula \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \):

\( \sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

Exam Tip: Remember that \( \sec(x) = \frac{1}{\cos(x)} \), so use cosine values to determine secant values when needed.

Question 5. Find the principal value of each of the following:
\( \csc^{-1}(-\sqrt{2}) \)
Answer: Using the formula \( \csc^{-1}(-x) = -\csc^{-1}(x) \):

\( \csc^{-1}(-\sqrt{2}) = -\csc^{-1}(\sqrt{2}) = -\frac{\pi}{4} \)

Alternatively, since \( \csc \) is negative in the third quadrant, the angle can also be expressed as \( \pi + \frac{\pi}{4} = \frac{5\pi}{4} \), but the principal value is \( -\frac{\pi}{4} \).

Exam Tip: Pay attention to the range of principal values for each inverse function - they differ and affect which answer is correct.

Question 6. Find the principal value of each of the following:
\( \cot^{-1}(-1) \)
Answer: Using the formula \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \):

\( \cot^{-1}(-1) = \pi - \cot^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

Exam Tip: Recognize that \( \cot\left(\frac{\pi}{4}\right) = 1 \), making this a straightforward application of the negative argument formula.

Question 7. Find the principal value of each of the following:
\( \tan^{-1}(-\sqrt{3}) \)
Answer: Using the formula \( \tan^{-1}(-x) = -\tan^{-1}(x) \):

\( \tan^{-1}(-\sqrt{3}) = -\tan^{-1}(\sqrt{3}) = -\frac{\pi}{3} \)

Exam Tip: Recognize that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \) is a standard value that appears frequently in these problems.

Question 8. Find the principal value of each of the following:
\( \sec^{-1}\left(\frac{-2}{\sqrt{3}}\right) \)
Answer: Using the formula \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \):

\( \sec^{-1}\left(\frac{-2}{\sqrt{3}}\right) = \pi - \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Exam Tip: When working with secant, convert to cosine values if needed: \( \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \).

Question 9. Find the principal value of each of the following:
\( \csc^{-1}(2) \)
Answer: Finding the principal value directly by identifying which angle has cosecant equal to 2:

\( \csc^{-1}(2) = \frac{\pi}{6} \)

This is because \( \csc\left(\frac{\pi}{6}\right) = 2 \).

Exam Tip: Recall that \( \csc(x) = \frac{1}{\sin(x)} \), so if \( \csc(x) = 2 \), then \( \sin(x) = \frac{1}{2} \).

Question 10. Find the principal value of each of the following:
\( \sin^{-1}\left(\sin\frac{2\pi}{3}\right) \)
Answer: The angle \( \frac{2\pi}{3} \) is in the second quadrant. Using the identity \( \sin\left(\pi - x\right) = \sin(x) \):

\( \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) \)

Therefore:

\( \sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3} \)

Exam Tip: When angles fall outside the principal value range, reduce them using complementary or supplementary angle identities before applying the inverse function.

Question 11. Find the principal value of each of the following:
\( \tan^{-1}\left(\tan\frac{3\pi}{4}\right) \)
Answer: The angle \( \frac{3\pi}{4} \) is in the second quadrant. Using the identity \( \tan\left(\pi - x\right) = -\tan(x) \), since tangent is negative in the second quadrant:

\( \tan\left(\frac{3\pi}{4}\right) = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) \)

Therefore:

\( \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}\left(-\tan\frac{\pi}{4}\right) = -\frac{\pi}{4} \)

Exam Tip: Angles in different quadrants have different signs for tangent. Always use the appropriate angle reduction formula based on the quadrant.

Question 12. Find the principal value of each of the following:
\( \cos^{-1}\left(\cos\frac{7\pi}{6}\right) \)
Answer: The angle \( \frac{7\pi}{6} \) is in the third quadrant. Using the identity \( \cos\left(2\pi - x\right) = \cos(x) \):

\( \cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{5\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \)

Therefore:

\( \cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \cos^{-1}\left(\cos\frac{5\pi}{6}\right) = \frac{5\pi}{6} \)

Exam Tip: For cosine, reduce angles to the interval \( [0, \pi] \) using appropriate periodicity formulas before finding the inverse.

Question 13. Find the principal value of each of the following:
\( \cos^{-1}\left(\cos\frac{13\pi}{6}\right) \)
Answer: Since cosine has a period of \( 2\pi \), reduce the angle:

\( \cos\left(\frac{13\pi}{6}\right) = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) \)

Therefore:

\( \cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\cos\frac{\pi}{6}\right) = \frac{\pi}{6} \)

Exam Tip: Always remove full rotations of \( 2\pi \) from angles to bring them into the standard range for inverse trigonometric functions.

Question 14. Find the principal value of each of the following:
\( \tan^{-1}\left(\tan\frac{7\pi}{6}\right) \)
Answer: Since tangent has a period of \( \pi \), reduce the angle:

\( \tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) \)

Therefore:

\( \tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left(\tan\frac{\pi}{6}\right) = \frac{\pi}{6} \)

Exam Tip: Remember that tangent repeats every \( \pi \) radians, making it easier to reduce angles to the principal value range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

Question 15. Find the principal value of each of the following:
\( \tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) \)
Answer: First, evaluate each inverse function separately.

For the first term: \( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \)

For the second term, use the formula \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \):

\( \cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Therefore:

\( \tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6} = \frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2} \)

Exam Tip: Evaluate each component separately before combining them - this avoids mixing up the formulas and reduces careless errors.

Question 16. Find the principal value of each of the following:
\( \sin\left[\frac{\pi}{3} - \sin^{-1}\left(\frac{-1}{2}\right)\right] \)
Answer: First, evaluate the inverse function using the formula \( \sin^{-1}(-x) = -\sin^{-1}(x) \):

\( \sin^{-1}\left(\frac{-1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6} \)

Substituting this value:

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

Exam Tip: Pay careful attention to signs when subtracting negative values - a double negative becomes positive.

Question 17. Find the principal value of each of the following:
\( \cot(\tan^{-1}x + \cot^{-1}x) \)
Answer: Using the fundamental formula \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \):

\( \cot(\tan^{-1}x + \cot^{-1}x) = \cot\left(\frac{\pi}{2}\right) = 0 \)

Exam Tip: This key relationship - that the sum of inverse tangent and inverse cotangent equals \( \frac{\pi}{2} \) - appears frequently. Memorizing it saves time.

Question 18. Find the principal value of each of the following:
\( \csc(\sin^{-1}x + \cos^{-1}x) \)
Answer: Using the fundamental formula \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \):

\( \csc(\sin^{-1}x + \cos^{-1}x) = \csc\left(\frac{\pi}{2}\right) = 1 \)

Exam Tip: Another essential identity - the sum of inverse sine and inverse cosine is always \( \frac{\pi}{2} \). This simplification is crucial for many problems.

Question 19. Find the principal value of each of the following:
\( \sin(\sec^{-1}x + \csc^{-1}x) \)
Answer: Using the fundamental formula \( \sec^{-1}x + \csc^{-1}x = \frac{\pi}{2} \):

\( \sin(\sec^{-1}x + \csc^{-1}x) = \sin\left(\frac{\pi}{2}\right) = 1 \)

Exam Tip: The complementary relationship between inverse secant and inverse cosecant mirrors that of sine and cosine - their sum is \( \frac{\pi}{2} \).

Question 20. Find the principal value of each of the following:
\( \cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2} \)
Answer: Evaluate each inverse function:

\( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \)

\( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \)

Substituting these values:

\( \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \)

Exam Tip: Always compute standard inverse values first, then perform arithmetic operations carefully, combining fractions only after conversion to a common denominator.

Question 21. Find the principal value of each of the following:
\( \tan^{-1}1 + \cos^{-1}\left(\frac{-1}{2}\right) + \sin^{-1}\left(\frac{-1}{2}\right) \)
Answer: Using the appropriate formulas for negative arguments:

\( \tan^{-1}1 = \frac{\pi}{4} \)

\( \cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

\( \sin^{-1}\left(\frac{-1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6} \)

Combining all three:

\( \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} \)

Exam Tip: When adding multiple inverse trigonometric values, convert all fractions to a common denominator before combining - this prevents arithmetic mistakes.

Question 22. Find the principal value of each of the following:
\( \sin^{-1}\left(\sin\frac{3\pi}{5}\right) \)
Answer: The angle \( \frac{3\pi}{5} \) is in the second quadrant. Using the identity \( \sin(\pi - x) = \sin(x) \):

\( \sin\left(\frac{3\pi}{5}\right) = \sin\left(\pi - \frac{2\pi}{5}\right) = \sin\left(\frac{2\pi}{5}\right) \)

Therefore:

\( \sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\frac{2\pi}{5}\right) = \frac{2\pi}{5} \)

Exam Tip: Angles in the second quadrant can be rewritten using \( \sin(\pi - x) = \sin(x) \) to bring them into the principal value range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Exercise 4C

Question 1A. Prove that:
\( \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x, \quad x < 1 \)
Answer: To establish this identity, we use the angle addition formula for tangent.

Formula Used: \( \tan\left(\frac{\pi}{4} + A\right) = \frac{1 + \tan A}{1 - \tan A} \)

Proof:

Consider the left-hand side: \( \tan^{-1}\left(\frac{1+x}{1-x}\right) \)

Let \( x = \tan A \) ... (2)

Substituting (2) into the LHS:

\( \text{LHS} = \tan^{-1}\left(\frac{1+\tan A}{1-\tan A}\right) \)

Using the formula above:

\( = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + A\right)\right) \)

\( = \frac{\pi}{4} + A \)

From (2), \( A = \tan^{-1}x \), therefore:

\( \frac{\pi}{4} + A = \frac{\pi}{4} + \tan^{-1}x \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: Angle addition formulas for inverse trigonometric functions are crucial. Always identify the pattern in the argument to match it with a known formula.

Question 1B. Prove that:
\( \tan^{-1}x + \cot^{-1}(x+1) = \tan^{-1}(x^2+x+1) \)
Answer: This proof uses the addition formula for inverse tangent functions.

Formula Used:
1) \( \cot^{-1}y = \tan^{-1}\frac{1}{y} \)

2) \( \tan^{-1}p + \tan^{-1}q = \tan^{-1}\left(\frac{p+q}{1-pq}\right) \)

Proof:

\( \text{LHS} = \tan^{-1}x + \cot^{-1}(x+1) \) ... (1)

Converting the cotangent inverse:

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

Using the addition formula:

\( = \tan^{-1}\left(\frac{x + \frac{1}{x+1}}{1 - x \cdot \frac{1}{x+1}}\right) \)

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

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

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

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: When dealing with cotangent inverses in addition formulas, convert them to tangent form immediately for consistency and easier manipulation.

Question 2. Prove that:
\( \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x, \quad |x| \leq \frac{1}{\sqrt{2}} \)
Answer: This identity stems from the double angle formula for sine.

Formula Used: \( \sin 2A = 2 \sin A \cos A \)

Proof:

\( \text{LHS} = \sin^{-1}(2x\sqrt{1-x^2}) \) ... (1)

Let \( x = \sin A \) ... (2)

Substituting (2) into (1):

\( \text{LHS} = \sin^{-1}(2\sin A\sqrt{1-\sin^2A}) \)

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

\( = \sin^{-1}(\sin 2A) \)

\( = 2A \)

From (2), \( A = \sin^{-1}x \), therefore:

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

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: Double angle formulas are essential for these types of proofs. Always substitute the inverse function with an angle variable to match the argument to a known identity.

Question 3A. Prove that:
\( \sin^{-1}(3x-4x^3) = 3\sin^{-1}x, \quad |x| \leq \frac{1}{2} \)
Answer: This identity uses the triple angle formula for sine.

Formula Used: \( \sin 3A = 3\sin A - 4\sin^3 A \)

Proof:

\( \text{LHS} = \sin^{-1}(3x-4x^3) \) ... (1)

Let \( x = \sin A \) ... (2)

Substituting (2) into (1):

\( \text{LHS} = \sin^{-1}(3\sin A - 4\sin^3 A) \)

\( = \sin^{-1}(\sin 3A) \)

\( = 3A \)

From (2), \( A = \sin^{-1}x \), therefore:

\( 3A = 3\sin^{-1}x \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: Triple angle formulas follow a similar pattern to double angle formulas. Memorizing these identity patterns helps solve these problems quickly and confidently.

Question 3B. Prove that:
\( \cos^{-1}(4x^3-3x) = 3\cos^{-1}x \)
Answer: This identity uses the triple angle formula for cosine.

Formula Used: \( \cos 3A = 4\cos^3 A - 3\cos A \)

Proof:

\( \text{LHS} = \cos^{-1}(4x^3-3x) \) ... (1)

Let \( x = \cos A \) ... (2)

Substituting (2) into (1):

\( \text{LHS} = \cos^{-1}(4\cos^3 A - 3\cos A) \)

\( = \cos^{-1}(\cos 3A) \)

\( = 3A \)

From (2), \( A = \cos^{-1}x \), therefore:

\( 3A = 3\cos^{-1}x \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: The cosine and sine triple angle formulas have different forms - pay careful attention to the signs and coefficients when applying them.

Question 3C. Prove that:
\( \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) = 3\tan^{-1}x, \quad |x| < \frac{1}{\sqrt{3}} \)
Answer: This identity stems from the triple angle formula for tangent.

Formula Used: \( \tan 3A = \frac{3\tan A - \tan^3 A}{1-3\tan^2 A} \)

Proof:

\( \text{LHS} = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) \) ... (1)

Let \( x = \tan A \) ... (2)

Substituting (2) into (1):

\( \text{LHS} = \tan^{-1}\left(\frac{3\tan A - \tan^3 A}{1-3\tan^2 A}\right) \)

\( = \tan^{-1}(\tan 3A) \)

\( = 3A \)

From (2), \( A = \tan^{-1}x \), therefore:

\( 3A = 3\tan^{-1}x \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: The tangent triple angle formula is more complex with both numerator and denominator involving multiple terms. Write it out carefully to avoid algebraic mistakes during substitution.

Question 3D. Prove that:
\( \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) \)
Answer: This combines the addition formula for inverse tangent with the double angle pattern.

Formula Used: \( \tan^{-1}p + \tan^{-1}q = \tan^{-1}\left(\frac{p+q}{1-pq}\right) \)

Proof:

\( \text{LHS} = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) ... (1)

Using the addition formula with \( p = x \) and \( q = \frac{2x}{1-x^2} \):

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

\( = \tan^{-1}\left(\frac{\frac{3x-x^3}{1-x^2}}{\frac{1-3x^2}{1-x^2}}\right) \)

\( = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) \)

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: When using the addition formula, carefully align numerators and denominators, and always find common denominators before combining fractions.

Question 4A. Prove that:
\( \cos^{-1}(1-2x^2) = 2\sin^{-1}x \)
Answer: This uses the double angle formula for cosine expressed in terms of sine.

Formula Used: \( \cos 2A = 1 - 2\sin^2 A \)

Proof:

\( \text{LHS} = \cos^{-1}(1-2x^2) \) ... (1)

Let \( x = \sin A \) ... (2)

Substituting (2) into (1):

\( \text{LHS} = \cos^{-1}(1-2\sin^2 A) \)

\( = \cos^{-1}(\cos 2A) \)

\( = 2A \)

From (2), \( A = \sin^{-1}x \), therefore:

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

\( = \text{RHS} \)

Therefore, LHS = RHS. Hence proved.

Exam Tip: This formula is particularly useful because it connects inverse cosine directly to inverse sine through the double angle relationship - a powerful tool in many problems.

Question 1. Prove that: \( \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x \)
Answer: Let \( x = \sin A \). Then \( \sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(2\sin A\sqrt{1-\sin^2 A}) = \sin^{-1}(2\sin A \cos A) = \sin^{-1}(\sin 2A) = 2A = 2\sin^{-1}x \). Thus, LHS = RHS and the result is established.
In simple words: Substitute \( x = \sin A \), apply the double angle formula \( \sin 2A = 2\sin A\cos A \), and simplify to show both sides are equal.

Exam Tip: Always express the given expression in terms of a single angle using substitution, then apply the relevant double angle formula to match the required form.

Question 2. Prove that: \( \cos^{-1}(2x^2-1) = 2\cos^{-1}x \)
Answer: Let \( x = \cos A \). Substituting this into the expression: \( \cos^{-1}(2x^2-1) = \cos^{-1}(2\cos^2 A - 1) = \cos^{-1}(\cos 2A) = 2A = 2\cos^{-1}x \). Therefore, LHS equals RHS and the identity is verified.
In simple words: Set \( x = \cos A \), use the identity \( \cos 2A = 2\cos^2 A - 1 \), and simplify to obtain the required equality.

Exam Tip: The key is recognizing the double angle formula within the argument and matching it to the appropriate inverse function identity.

Question 3. Prove that: \( \sec^{-1}\left(\frac{1}{2x^2-1}\right) = 2\cos^{-1}x \)
Answer: Since \( \sec^{-1}\left(\frac{1}{y}\right) = \cos^{-1}(y) \), we have \( \sec^{-1}\left(\frac{1}{2x^2-1}\right) = \cos^{-1}(2x^2-1) \). Let \( x = \cos A \). Then \( \cos^{-1}(2\cos^2 A - 1) = \cos^{-1}(\cos 2A) = 2A = 2\cos^{-1}x \). Hence, LHS = RHS.
In simple words: Convert the secant inverse to cosine inverse, substitute \( x = \cos A \), apply the double angle formula, and verify the equality.

Exam Tip: Remember that \( \sec^{-1}(z) = \cos^{-1}(1/z) \); this relationship simplifies expressions involving secant inverse functions.

Question 4. Prove that: \( \cot^{-1}\left(\sqrt{1+x^2}-x\right) = \frac{\pi}{2} - \frac{1}{2}\cot^{-1}x \)
Answer: Let \( x = \cot A \). Then \( \cot^{-1}(\sqrt{1+\cot^2 A} - \cot A) = \cot^{-1}(\text{cosec } A - \cot A) = \cot^{-1}\left(\frac{1-\cos A}{\sin A}\right) = \cot^{-1}\left(\frac{2\sin^2(A/2)}{2\sin(A/2)\cos(A/2)}\right) = \cot^{-1}(\tan(A/2)) = \frac{\pi}{2} - \tan^{-1}(\tan(A/2)) = \frac{\pi}{2} - \frac{A}{2} = \frac{\pi}{2} - \frac{1}{2}\cot^{-1}x \). Thus, LHS = RHS.
In simple words: Substitute \( x = \cot A \), simplify using half-angle identities, and recognize that \( \cot^{-1}(\tan(\theta)) = \frac{\pi}{2} - \theta \).

Exam Tip: Expressions involving square roots of the form \( \sqrt{1+x^2} \) often benefit from trigonometric substitution followed by half-angle simplifications.

Question 5. Prove that: \( \tan^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}\right) = \tan^{-1}\sqrt{x} + \tan^{-1}\sqrt{y} \)
Answer: Using the identity \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) when \( AB < 1 \), set \( A = \sqrt{x} \) and \( B = \sqrt{y} \). Then \( \tan^{-1}\sqrt{x} + \tan^{-1}\sqrt{y} = \tan^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x}\cdot\sqrt{y}}\right) = \tan^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}\right) \). Therefore, the result is proved.
In simple words: Apply the sum formula for inverse tangent with \( A = \sqrt{x} \) and \( B = \sqrt{y} \) to directly obtain the required identity.

Exam Tip: Always check that the product of the arguments is less than 1 before applying the sum formula for inverse tangent to ensure validity.

Question 6. Prove that: \( \tan^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3/2}}\right) = \tan^{-1}x + \tan^{-1}\sqrt{x} \)
Answer: Using the addition formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) with \( A = x \) and \( B = \sqrt{x} \), we obtain \( \tan^{-1}x + \tan^{-1}\sqrt{x} = \tan^{-1}\left(\frac{x+\sqrt{x}}{1-x\cdot\sqrt{x}}\right) = \tan^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3/2}}\right) \). Thus, LHS equals RHS.
In simple words: Use the inverse tangent addition formula where the two terms are \( x \) and \( \sqrt{x} \), noting that \( x \cdot \sqrt{x} = x^{3/2} \).

Exam Tip: When dealing with expressions like \( x^{3/2} \), recognize it as the product \( x \cdot x^{1/2} \), which often arises naturally from the addition formula.

Question 7. Prove that: \( \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) = \frac{x}{2} \)
Answer: Using the identities \( \sin x = 2\sin(x/2)\cos(x/2) \) and \( 1 + \cos x = 2\cos^2(x/2) \), we have \( \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) = \tan^{-1}\left(\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}\right) = \tan^{-1}\left(\frac{\sin(x/2)}{\cos(x/2)}\right) = \tan^{-1}(\tan(x/2)) = \frac{x}{2} \). Therefore, the identity is established.
In simple words: Apply half-angle formulas to the numerator and denominator, simplify to get the tangent of half the angle, and then apply the inverse tangent.

Exam Tip: Half-angle identities are powerful tools for simplifying trigonometric expressions; always look for patterns like \( 2\sin\theta\cos\theta \) or \( 2\cos^2\theta \).

Question 8. Prove that: \( \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{2}{11} = \tan^{-1}\frac{3}{4} \)
Answer: Using the formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) where \( xy < 1 \), we compute: LHS \( = \tan^{-1}\left(\frac{\frac{1}{2}+\frac{2}{11}}{1-\frac{1}{2}\cdot\frac{2}{11}}\right) = \tan^{-1}\left(\frac{\frac{11+4}{22}}{1-\frac{1}{11}}\right) = \tan^{-1}\left(\frac{\frac{15}{22}}{\frac{10}{11}}\right) = \tan^{-1}\left(\frac{15}{22}\cdot\frac{11}{10}\right) = \tan^{-1}\left(\frac{3}{4}\right) \) = RHS. Hence proved.
In simple words: Apply the sum formula for inverse tangent, compute the numerator and denominator carefully, simplify the resulting fraction, and verify it equals \( \tan^{-1}(3/4) \).

Exam Tip: When adding inverse tangent values, always verify the condition \( xy < 1 \) first, then substitute into the formula and simplify step-by-step without skipping arithmetic.

Question 9. Prove that: \( \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} = \tan^{-1}\frac{1}{2} \)
Answer: Applying the addition formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) with \( x = \frac{2}{11} \) and \( y = \frac{7}{24} \): LHS \( = \tan^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\cdot\frac{7}{24}}\right) = \tan^{-1}\left(\frac{\frac{48+77}{264}}{1-\frac{7}{132}}\right) = \tan^{-1}\left(\frac{\frac{125}{264}}{\frac{125}{132}}\right) = \tan^{-1}\left(\frac{125}{264}\cdot\frac{132}{125}\right) = \tan^{-1}\left(\frac{1}{2}\right) \) = RHS. Hence proved.
In simple words: Use the inverse tangent addition formula, compute the numerator by finding a common denominator, calculate the denominator using the product, then simplify the complex fraction.

Exam Tip: Fractions within fractions require careful handling; cross-multiply or multiply by the reciprocal to simplify, and always double-check your arithmetic.

Question 10. Prove that: \( \tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{2} \)
Answer: First, compute \( \tan^{-1}1 + \tan^{-1}\frac{1}{2} = \tan^{-1}\left(\frac{1+\frac{1}{2}}{1-1\cdot\frac{1}{2}}\right) = \tan^{-1}\left(\frac{\frac{3}{2}}{\frac{1}{2}}\right) = \tan^{-1}3 \). Next, apply the formula for \( \tan^{-1}3 + \tan^{-1}\frac{1}{3} \): since the product is \( 3 \times \frac{1}{3} = 1 \), we use \( \tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2} \) for \( x > 0 \). Therefore, LHS \( = \tan^{-1}3 + \tan^{-1}\frac{1}{3} = \frac{\pi}{2} \) = RHS. Hence proved.
In simple words: Combine the first two inverse tangent terms using the addition formula, recognize that the result is \( \tan^{-1}3 \), then use the fact that \( \tan^{-1}x + \tan^{-1}(1/x) = \pi/2 \) for positive \( x \).

Exam Tip: Know the special identity \( \tan^{-1}x + \tan^{-1}(1/x) = \frac{\pi}{2} \) for \( x > 0 \); it frequently simplifies these proofs significantly.

Question 11. Prove that: \( 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4} \)
Answer: Using the double angle formula, \( 2\tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{2\cdot\frac{1}{3}}{1-(\frac{1}{3})^2}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) = \tan^{-1}\left(\frac{3}{4}\right) \). Now apply the addition formula: \( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7} = \tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\cdot\frac{1}{7}}\right) = \tan^{-1}\left(\frac{\frac{21+4}{28}}{1-\frac{3}{28}}\right) = \tan^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) = \tan^{-1}1 = \frac{\pi}{4} \). Hence proved.
In simple words: First, apply the double angle formula to \( 2\tan^{-1}(1/3) \) to get \( \tan^{-1}(3/4) \), then add \( \tan^{-1}(1/7) \) using the sum formula to arrive at \( \pi/4 \).

Exam Tip: The double angle formula for inverse tangent is \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \); use it when coefficients are doubled in the problem.

Question 12. Prove that: \( \tan^{-1}2 - \tan^{-1}1 = \tan^{-1}\frac{1}{3} \)
Answer: Applying the subtraction formula \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \) where \( xy > -1 \): LHS \( = \tan^{-1}\left(\frac{2-1}{1+2\cdot1}\right) = \tan^{-1}\left(\frac{1}{3}\right) \) = RHS. Therefore, the identity is proved.
In simple words: Use the inverse tangent subtraction formula with \( x = 2 \) and \( y = 1 \) to directly obtain \( \tan^{-1}(1/3) \).

Exam Tip: The subtraction formula for inverse tangent mirrors the addition formula; remember the denominator changes to \( 1 + xy \) instead of \( 1 - xy \).

Question 13. Prove that: \( \tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi \)
Answer: First, combine \( \tan^{-1}1 + \tan^{-1}2 \): Since the product \( 1 \times 2 = 2 > 1 \), we use \( \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) for \( xy > 1 \). This gives \( \tan^{-1}1 + \tan^{-1}2 = \pi + \tan^{-1}\left(\frac{1+2}{1-2}\right) = \pi + \tan^{-1}(-3) = \pi - \tan^{-1}3 \). Now add \( \tan^{-1}3 \): LHS \( = \pi - \tan^{-1}3 + \tan^{-1}3 = \pi \) = RHS. Hence proved.
In simple words: When the product of two arguments exceeds 1, apply the modified addition formula involving \( \pi \), simplify to get \( \pi - \tan^{-1}3 \), then add \( \tan^{-1}3 \) to obtain \( \pi \).

Exam Tip: Always check the product \( xy \) before applying the addition formula; when \( xy > 1 \), add or subtract \( \pi \) from the result as needed.

Question 14. Prove that: \( \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4} \)
Answer: Combine \( \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} \): LHS for this pair \( = \tan^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\cdot\frac{1}{8}}\right) = \tan^{-1}\left(\frac{\frac{8+5}{40}}{1-\frac{1}{40}}\right) = \tan^{-1}\left(\frac{\frac{13}{40}}{\frac{39}{40}}\right) = \tan^{-1}\left(\frac{1}{3}\right) \). Now add \( \tan^{-1}\frac{1}{2} \): \( \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}1 = \frac{\pi}{4} \). Hence proved.
In simple words: First combine the smaller fractions \( 1/5 \) and \( 1/8 \) to get \( \tan^{-1}(1/3) \), then add \( \tan^{-1}(1/2) \) to arrive at \( \pi/4 \).

Exam Tip: When adding three or more inverse tangent terms, strategically group pairs that simplify nicely, then combine the result with the remaining term.

Question 15. Prove that: \( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \)
Answer: Using the formula \( \cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) \): LHS \( = \cos^{-1}\left(\frac{4}{5}\cdot\frac{12}{13} - \sqrt{1-\frac{16}{25}}\sqrt{1-\frac{144}{169}}\right) = \cos^{-1}\left(\frac{48}{65} - \sqrt{\frac{9}{25}}\sqrt{\frac{25}{169}}\right) = \cos^{-1}\left(\frac{48}{65} - \frac{3}{5}\cdot\frac{5}{13}\right) = \cos^{-1}\left(\frac{48}{65} - \frac{15}{65}\right) = \cos^{-1}\left(\frac{33}{65}\right) \) = RHS. Hence proved.
In simple words: Apply the addition formula for inverse cosine, compute each square root term from the given values, multiply and subtract to obtain the final result.

Exam Tip: For the cosine addition formula, memorize: \( \cos^{-1}x + \cos^{-1}y = \cos^{-1}\left(xy - \sqrt{(1-x^2)(1-y^2)}\right) \); compute the radical terms carefully.

Question 16. Prove that: \( \sin^{-1}\frac{1}{\sqrt{5}} + \sin^{-1}\frac{2}{\sqrt{5}} = \frac{\pi}{2} \)
Answer: Using the formula \( \sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2}) \): LHS \( = \sin^{-1}\left(\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}} + \frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\sqrt{\frac{1}{5}} + \frac{2}{\sqrt{5}}\sqrt{\frac{4}{5}}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}}\cdot\frac{2}{\sqrt{5}}\right) = \sin^{-1}\left(\frac{1}{5} + \frac{4}{5}\right) = \sin^{-1}1 = \frac{\pi}{2} \). Hence proved.
In simple words: Apply the inverse sine addition formula, calculate each square root term, multiply and add to get 1, then apply \( \sin^{-1}1 = \pi/2 \).

Exam Tip: The sine addition formula is \( \sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2}) \); when the sum equals 1, the result is \( \pi/2 \).

Question 17. Prove that: \( \cos^{-1}\frac{3}{5} + \sin^{-1}\frac{12}{13} = \sin^{-1}\frac{56}{65} \)
Answer: Convert \( \cos^{-1}\frac{3}{5} \) to sine form using a right triangle: if \( \cos\theta = \frac{3}{5} \), then \( \sin\theta = \frac{4}{5} \), so \( \cos^{-1}\frac{3}{5} = \sin^{-1}\frac{4}{5} \). Now apply the sine addition formula: \( \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{12}{13} = \sin^{-1}\left(\frac{4}{5}\sqrt{1-\frac{144}{169}} + \frac{12}{13}\sqrt{1-\frac{16}{25}}\right) = \sin^{-1}\left(\frac{4}{5}\cdot\frac{5}{13} + \frac{12}{13}\cdot\frac{3}{5}\right) = \sin^{-1}\left(\frac{20}{65} + \frac{36}{65}\right) = \sin^{-1}\frac{56}{65} \). Hence proved.
In simple words: Convert the cosine inverse to sine form using the complementary angle, apply the sine addition formula, and simplify the resulting fractions.

Exam Tip: When mixing sine and cosine inverse functions, convert one to the other using a right triangle or the complementary angle relationship, then apply a consistent addition formula.

Question 18. Prove that: \( \tan^{-1}\frac{1}{3} + \sec^{-1}\frac{\sqrt{5}}{2} = \frac{\pi}{4} \)
Answer: Convert \( \sec^{-1}\frac{\sqrt{5}}{2} \) to tangent form. If \( \sec\alpha = \frac{\sqrt{5}}{2} \), then \( \cos\alpha = \frac{2}{\sqrt{5}} \), and \( \tan\alpha = \frac{1}{2} \) (from a right triangle). Thus, \( \sec^{-1}\frac{\sqrt{5}}{2} = \tan^{-1}\frac{1}{2} \). Now apply the tangent addition formula: \( \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2} = \tan^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}1 = \frac{\pi}{4} \). Hence proved.
In simple words: Convert the secant inverse to tangent inverse using a right triangle, then apply the tangent addition formula to obtain \( \pi/4 \).

Exam Tip: Always convert non-standard inverse functions like \( \sec^{-1} \), \( \csc^{-1} \), or \( \cot^{-1} \) to standard ones before applying addition/subtraction formulas.

Question 7G. Prove that: \( 2\sin^{-1}\frac{3}{5} - \tan^{-1}\frac{17}{31} = \frac{\pi}{4} \)
Answer: To establish the result, the following formulas are used: (1) \( 2\sin^{-1}x = \sin^{-1}(2x\sqrt{1-x^2}) \), (2) \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) where \( xy < 1 \).

Working from the left side: \( 2\sin^{-1}\frac{3}{5} = \sin^{-1}\left(2 \times \frac{3}{5} \times \sqrt{1-\left(\frac{3}{5}\right)^2}\right) = \sin^{-1}\left(\frac{6}{5} \times \frac{4}{5}\right) = \sin^{-1}\frac{24}{25} \) ... (1)

Substituting (1) into the original equation: \( \sin^{-1}\frac{24}{25} - \tan^{-1}\frac{17}{31} \) ... (2)

Let \( \sin\theta = \frac{24}{25} \), so \( \theta = \sin^{-1}\frac{24}{25} \) ... (3)

From the right triangle with hypotenuse 25 and opposite side 24, the adjacent side is 7, giving \( \tan\theta = \frac{24}{7} \) ... (4)

Therefore: \( \tan^{-1}\frac{24}{25} - \tan^{-1}\frac{17}{31} = \tan^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times\frac{17}{31}}\right) = \tan^{-1}\left(\frac{744-119}{217+408}\right) = \tan^{-1}\left(\frac{625}{625}\right) = \tan^{-1}1 = \frac{\pi}{4}\)
In simple words: Use the doubling formula to convert the first term, convert everything to inverse tangent form, apply the addition formula, and simplify to get \( \tan^{-1}1 = \frac{\pi}{4} \).

Exam Tip: Always identify which inverse trigonometric formulas apply and work systematically to convert all terms to the same function before combining them.

Question 8A. Solve for x: \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} \)
Answer: The goal is to find the value of x.

Using the formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) where \( xy < 1 \):

\( \tan^{-1}\left(\frac{(x+1)+(x-1)}{1-((x+1)(x-1))}\right) = \tan^{-1}\left(\frac{2x}{1-(x^2-1)}\right) = \tan^{-1}\left(\frac{2x}{2-x^2}\right) = \tan^{-1}\frac{8}{31} \)

Taking tangent on both sides: \( \frac{2x}{2-x^2} = \frac{8}{31} \)

This leads to: \( 62x = 16 - 8x^2 \)

\( 8x^2 + 62x - 16 = 0 \)

\( 4x^2 + 31x - 8 = 0 \)

\( 4x^2 + 32x - x - 8 = 0 \)

\( 4x(x+8) - 1(x+8) = 0 \)

\( (4x-1)(x+8) = 0 \)

Therefore, \( x = \frac{1}{4} \) or \( x = -8 \) are the solutions.
In simple words: Apply the formula for adding two inverse tangents, equate the resulting expression to the right side, and solve the resulting quadratic equation by factoring.

Exam Tip: Always check that the condition \( xy < 1 \) is satisfied before using the addition formula, and verify both solutions in the original equation.

Question 8B. Solve for x: \( \tan^{-1}(2+x) + \tan^{-1}(2-x) = \tan^{-1}\frac{2}{3} \)
Answer: The objective is to find the value of x.

Using the formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) where \( xy < 1 \):

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

Taking tangent on both sides: \( \frac{4}{x^2-3} = \frac{2}{3} \)

This simplifies to: \( 12 = 2x^2 - 6 \)

\( 2x^2 = 18 \)

\( x^2 = 9 \)

\( x = 3 \) or \( x = -3 \)

Therefore, \( x = \pm 3 \) are the required solutions.
In simple words: Combine the two inverse tangent terms using the addition formula, simplify the resulting fraction, and then solve the equation by cross-multiplying and isolating x.

Exam Tip: Simplify the denominator carefully when \( xy \) involves difference of squares, as this often leads to a clean quadratic.

Question 8C. Solve for x: \( \cos(\sin^{-1}x) = \frac{1}{9} \)
Answer: The goal is to find the value of x.

Let \( \sin\theta = x \), therefore \( \theta = \sin^{-1}x \) ... (1)

From a right triangle where the opposite side is x and the hypotenuse is 1, the adjacent side is \( \sqrt{1-x^2} \), giving \( \cos\theta = \sqrt{1-x^2} \) ... (2)

From (1) and (2): \( \sin^{-1}x = \cos^{-1}\sqrt{1-x^2} \) ... (3)

Substituting (3) into the original equation: \( \cos(\cos^{-1}\sqrt{1-x^2}) = \sqrt{1-x^2} = \frac{1}{9} \)

Squaring both sides: \( 1 - x^2 = \frac{1}{81} \)

\( x^2 = 1 - \frac{1}{81} = \frac{80}{81} \)

\( x = \pm\frac{4\sqrt{5}}{9} \)

Therefore, \( x = \pm\frac{4\sqrt{5}}{9} \) are the required solutions.
In simple words: Express \( \cos(\sin^{-1}x) \) in terms of x using a right triangle diagram, set this equal to the given value, and solve by squaring.

Exam Tip: Always construct a right triangle to visualize the relationship between inverse trigonometric functions and identify adjacent and opposite sides correctly.

Question 8D. Solve for x: \( \cos(2\sin^{-1}x) = \frac{1}{9} \)
Answer: The objective is to find the value of x.

Using the formula \( 2\sin^{-1}x = \sin^{-1}(2x\sqrt{1-x^2}) \):

Let \( \theta = \sin^{-1}x \), so \( x = \sin\theta \) ... (1)

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

Substituting into the given equation: \( 1 - 2\sin^2\theta = \frac{1}{9} \)

\( 2\sin^2\theta = \frac{8}{9} \)

\( \sin^2\theta = \frac{4}{9} \)

Substituting in (1): \( x^2 = \frac{4}{9} \)

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

Therefore, \( x = \pm\frac{2}{3} \) are the required solutions.
In simple words: Express \( \cos(2\sin^{-1}x) \) using the double angle formula, substitute the variable, simplify the resulting equation, and solve for x.

Exam Tip: The double angle formula \( \cos(2\theta) = 1 - 2\sin^2\theta \) is most efficient here; remember to substitute back to the original variable at the end.

Question 8E. Solve for x: \( \sin^{-1}\frac{8}{x} + \sin^{-1}\frac{15}{x} = \frac{\pi}{2} \)
Answer: The goal is to find the value of x.

We use the fact that \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \)

Let \( \sin^{-1}\frac{8}{x} = P \), so \( \sin P = \frac{8}{x} \)

Therefore, \( \cos P = \frac{\sqrt{x^2-64}}{x} \) and \( P = \cos^{-1}\frac{\sqrt{x^2-64}}{x} \)

From the given equation: \( \cos^{-1}\frac{\sqrt{x^2-64}}{x} + \sin^{-1}\frac{15}{x} = \frac{\pi}{2} \)

Therefore, \( \frac{\sqrt{x^2-64}}{x} = \frac{15}{x} \)

\( \sqrt{x^2-64} = 15 \)

Squaring both sides: \( x^2 - 64 = 225 \)

\( x^2 = 289 \)

\( x = \pm 17 \)

Therefore, \( x = \pm 17 \) are the required solutions.
In simple words: Use the complementary angle relationship for inverse sines, extract the cosine of the first angle from a right triangle, equate and solve the resulting equation.

Exam Tip: Recognize that \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) allows you to convert one term into the complementary form, simplifying the equation significantly.

Question 9A. Solve for x: \( \cos(\sin^{-1}x) = \frac{1}{2} \)
Answer: The objective is to find the value of x.

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

Let \( \sin\theta = x \), therefore \( \cos\theta = \sqrt{1-x^2} \)

From the given equation: \( \sqrt{1-x^2} = \frac{1}{2} \)

Squaring both sides: \( 1 - x^2 = \frac{1}{4} \)

\( x^2 = \frac{3}{4} \)

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

Therefore, \( x = \pm\frac{\sqrt{3}}{2} \) are the required solutions.
In simple words: Convert the expression using a right triangle to find \( \cos(\sin^{-1}x) = \sqrt{1-x^2} \), then set it equal to the given value and solve.

Exam Tip: The relationship \( \cos(\sin^{-1}x) = \sqrt{1-x^2} \) is fundamental - draw the triangle every time to remember it clearly.

Question 9B. Solve for x: \( \tan^{-1}x = \sin^{-1}\frac{1}{\sqrt{2}} \)
Answer: The goal is to find the value of x.

Given: \( \tan^{-1}x = \sin^{-1}\frac{1}{\sqrt{2}} \)

We know that \( \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \)

Therefore, \( \frac{\pi}{4} = \sin^{-1}\frac{1}{\sqrt{2}} \)

Substituting in the given equation: \( \tan^{-1}x = \frac{\pi}{4} \)

\( x = \tan\frac{\pi}{4} \)

\( x = 1 \)

Therefore, \( x = 1 \) is the required solution.
In simple words: Evaluate \( \sin^{-1}\frac{1}{\sqrt{2}} \) by recalling that sine of \( \frac{\pi}{4} \) equals this value, then find x as the tangent of \( \frac{\pi}{4} \).

Exam Tip: Memorize standard inverse function values like \( \sin^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} \) and \( \tan^{-1}1 = \frac{\pi}{4} \) to solve these problems quickly.

Question 9C. Solve for x: \( \sin^{-1}x - \cos^{-1}x = \frac{\pi}{6} \)
Answer: The objective is to find the value of x.

Given: \( \sin^{-1}x - \cos^{-1}x = \frac{\pi}{6} \)

We know that \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \)

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

Substituting in the given equation: \( \frac{\pi}{2} - \cos^{-1}x - \cos^{-1}x = \frac{\pi}{6} \)

Rearranging: \( 2\cos^{-1}x = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \)

\( \cos^{-1}x = \frac{\pi}{6} \)

\( x = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \)

Therefore, \( x = \frac{\sqrt{3}}{2} \) is the required solution.
In simple words: Use the complementary angle identity to express \( \sin^{-1}x \) in terms of \( \cos^{-1}x \), substitute, simplify, and solve.

Exam Tip: Always apply the complementary relationship \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) when both inverse sine and cosine appear in the same equation.

Exercise 4D. Question 1. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \sin^{-1}x \)
Answer: The principal value branch of \( \sin^{-1}x \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)


In simple words: The function \( \sin^{-1}x \) is defined for x values between -1 and 1, and its output ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). The curve passes through the origin and rises from bottom left to top right, increasing throughout.

Exam Tip: Always remember that the principal branch restricts the range to ensure the inverse function is one-to-one; for sine inverse, this means the output is always between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

Exercise 4D. Question 2. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \cos^{-1}x \)
Answer: The principal value branch of \( \cos^{-1}x \) is \( [0, \pi] \)


In simple words: The function \( \cos^{-1}x \) is defined for x values between -1 and 1, and its output ranges from 0 to \( \pi \). The curve starts at the top left and decreases monotonically to the bottom right.

Exam Tip: Note that \( \cos^{-1}x \) always decreases as x increases, and it always outputs values in the range \( [0, \pi] \), which is the upper half of the unit circle.

Exercise 4D. Question 3. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \tan^{-1}x \)
Answer: The principal value branch of \( \tan^{-1}x \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)


In simple words: The function \( \tan^{-1}x \) accepts all real numbers as input and outputs values strictly between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \). The graph approaches but never touches the horizontal asymptotes at \( \pm\frac{\pi}{2} \).

Exam Tip: The open interval notation \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) indicates that the endpoint values are never actually achieved, though the function approaches them as x approaches \( \pm\infty \).

Exercise 4D. Question 4. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \cot^{-1}x \)
Answer: The principal value branch of \( \cot^{-1}x \) is \( (0, \pi) \)


In simple words: The function \( \cot^{-1}x \) is defined for all real numbers and outputs values strictly between 0 and \( \pi \). The graph steadily decreases from left to right and has horizontal asymptotes at \( y = 0 \) and \( y = \pi \).

Exam Tip: Remember that \( \cot^{-1}x \) is always decreasing and its range is the open interval \( (0, \pi) \), making it complementary to \( \tan^{-1}x \) in many ways.

Exercise 4D. Question 5. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \sec^{-1}x \)
Answer: The principal value branch of \( \sec^{-1}x \) is \( \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right] \)


In simple words: The function \( \sec^{-1}x \) is defined only when \( |x| \geq 1 \) and outputs values in two separate branches: from 0 to \( \frac{\pi}{2} \) (excluding \( \frac{\pi}{2} \)) and from \( \frac{\pi}{2} \) (excluding it) to \( \pi \). The graph has a vertical gap at \( y = \frac{\pi}{2} \).

Exam Tip: The domain of \( \sec^{-1}x \) is \( (-\infty, -1] \cup [1, \infty) \), and the range excludes \( \frac{\pi}{2} \) because secant is undefined there.

Exercise 4D. Question 6. Write down the interval for the principal-value branch of each of the following functions and draw its graph: \( \text{cosec}^{-1}x \)
Answer: The principal value branch of \( \text{cosec}^{-1}x \) is \( \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \)


In simple words: The function \( \text{cosec}^{-1}x \) is defined only when \( |x| \geq 1 \) and outputs values in two separate ranges: from \( -\frac{\pi}{2} \) to 0 (excluding 0) and from 0 (excluding it) to \( \frac{\pi}{2} \). The graph has a horizontal gap at \( y = 0 \).

Exam Tip: The domain of \( \text{cosec}^{-1}x \) is \( (-\infty, -1] \cup [1, \infty) \), and the range is split into two parts with the x-axis excluded because cosecant is undefined when sine equals zero.

Objective Questions. Question 1. Mark the tick against the correct answer in the following: The principal value of \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{5\pi}{6} \)
(c) \( \frac{7\pi}{6} \)
(d) None of the options
Answer: (a) \( \frac{\pi}{6} \)
In simple words: We need to find the angle whose cosine equals \( \frac{\sqrt{3}}{2} \). Since cosine of \( \frac{\pi}{6} \) equals \( \frac{\sqrt{3}}{2} \), and \( \frac{\pi}{6} \) lies in the principal range \( [0, \pi] \), the answer is \( \frac{\pi}{6} \).

Exam Tip: Always verify that the angle you find falls within the principal value range for the inverse function being used.

Objective Questions. Question 2. Mark the tick against the correct answer in the following: The principal value of \( \text{cosec}^{-1}(2) \) is
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{2\pi}{3} \)
(d) \( \frac{5\pi}{6} \)
Answer: (b) \( \frac{\pi}{6} \)
In simple words: We need to find the angle whose cosecant equals 2. Since cosecant is the reciprocal of sine, we need \( \sin\theta = \frac{1}{2} \). The angle in the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) with sine equal to \( \frac{1}{2} \) is \( \frac{\pi}{6} \).

Exam Tip: For inverse cosecant, convert to the equivalent sine equation first, then find the angle whose sine matches the given value within the allowed range.

Objective Questions. Question 3. Mark the tick against the correct answer in the following: The principal value of \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \) is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{6} \)
(d) None of the options
Answer: (c) \( \frac{\pi}{6} \)
In simple words: Find the angle whose cosine is \( \frac{\sqrt{3}}{2} \). In the principal range \( [0, \pi] \) for \( \cos^{-1} \), the angle whose cosine equals \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \).

Exam Tip: Memorize the standard values: \( \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \), \( \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( \cos\frac{\pi}{3} = \frac{1}{2} \) for quick reference.

Objective Questions. Question 4. Mark the tick against the correct answer in the following: The principal value of \( \cos^{-1}\left(\frac{-1}{\sqrt{2}}\right) \) is
(a) \( -\frac{\pi}{4} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{3\pi}{4} \)
(d) \( \frac{5\pi}{4} \)
Answer: (c) \( \frac{3\pi}{4} \)
In simple words: We seek the angle in \( [0, \pi] \) whose cosine equals \( \frac{-1}{\sqrt{2}} \). Since \( \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \), and cosine is negative in the second quadrant, the answer is \( \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

Exam Tip: When finding inverse cosine of a negative value, work from the corresponding positive angle and use the supplementary angle relationship within the range \( [0, \pi] \).

Objective Questions. Question 5. Mark the tick against the correct answer in the following: The principal value of \( \sin^{-1}\left(\frac{-1}{2}\right) \) is
(a) \( -\frac{\pi}{6} \)
(b) \( \frac{5\pi}{6} \)
(c) \( \frac{7\pi}{6} \)
(d) None of the options
Answer: (a) \( -\frac{\pi}{6} \)
In simple words: We need the angle in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) whose sine equals \( -\frac{1}{2} \). Since \( \sin\frac{\pi}{6} = \frac{1}{2} \), the angle with sine equal to \( -\frac{1}{2} \) is \( -\frac{\pi}{6} \).

Exam Tip: The principal range for inverse sine is symmetric around zero, so negative sine values give negative angles within \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Objective Questions. Question 6. Mark the tick against the correct answer in the following: The principal value of \( \cos^{-1}\left(\frac{-1}{2}\right) \) is
(a) \( -\frac{\pi}{3} \)
(b) \( \frac{2\pi}{3} \)
(c) \( \frac{4\pi}{3} \)
(d) \( \frac{\pi}{3} \)
Answer: (b) \( \frac{2\pi}{3} \)
In simple words: Find the angle in \( [0, \pi] \) whose cosine is \( -\frac{1}{2} \). Since \( \cos\frac{\pi}{3} = \frac{1}{2} \), and cosine is negative in the second quadrant, the answer is \( \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).

Exam Tip: For inverse cosine, the principal range \( [0, \pi] \) covers both the first and second quadrants, allowing both positive and negative cosine values.

Objective Questions. Question 7. Mark the tick against the correct answer in the following: The principal value of \( \tan^{-1}(-\sqrt{3}) \) is
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{4\pi}{3} \)
(c) \( -\frac{\pi}{3} \)
(d) None of the options
Answer: (c) \( -\frac{\pi}{3} \)
In simple words: Find the angle in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) whose tangent equals \( -\sqrt{3} \). Since \( \tan\frac{\pi}{3} = \sqrt{3} \), the angle with tangent equal to \( -\sqrt{3} \) is \( -\frac{\pi}{3} \).

Exam Tip: The principal range for inverse tangent is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), which includes both positive and negative angle values symmetrically.

Objective Questions. Question 8. Mark the tick against the correct answer in the following: The principal value of \( \cot^{-1}(-1) \) is
(a) \( -\frac{\pi}{4} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{3\pi}{4} \)
(d) \( \frac{5\pi}{4} \)
Answer: (c) \( \frac{3\pi}{4} \)
In simple words: Find the angle in \( (0, \pi) \) whose cotangent equals -1. Since \( \cot\frac{\pi}{4} = 1 \), and cotangent is negative in the second quadrant, the answer is \( \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

Exam Tip: The principal range for inverse cotangent is \( (0, \pi) \), which spans the first and second quadrants where cotangent takes all real values.

Objective Questions. Question 9. Mark the tick against the correct answer in the following: The principal value of \( \sec^{-1}\left(\frac{-2}{\sqrt{3}}\right) \) is
(a) \( \frac{\pi}{6} \)
(b) \( -\frac{\pi}{6} \)
(c) \( \frac{5\pi}{6} \)
(d) None of the options
Answer: (c) \( \frac{5\pi}{6} \)
In simple words: We need the angle in \( \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right] \) whose secant equals \( \frac{-2}{\sqrt{3}} \). Since \( \sec\frac{\pi}{6} = \frac{2}{\sqrt{3}} \), and secant is negative in the second quadrant, the answer is \( \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).

Exam Tip: For inverse secant of a negative value, find the corresponding positive angle and then locate it in the second quadrant of the principal range.

Objective Questions. Question 10. Mark the tick against the correct answer in the following: The principal value of \( \cot^{-1}(-\sqrt{3}) \) is
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{4\pi}{3} \)
(c) \( -\frac{\pi}{3} \)
(d) None of the options
Answer: (a) \( \frac{5\pi}{6} \)
In simple words: Find the angle in \( (0, \pi) \) whose cotangent equals \( -\sqrt{3} \). Since \( \cot\frac{\pi}{6} = \sqrt{3} \), and cotangent is negative in the second quadrant, the answer is \( \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).

Exam Tip: For inverse cotangent with a negative argument, use the supplementary angle relationship within the range \( (0, \pi) \).

Question 11. Mark the tick against the correct answer in the following: The value of \( \sin^{-1}\left(\sin\frac{2\pi}{3}\right) \) is
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{5\pi}{3} \)
(c) \( \frac{\pi}{3} \)
(d) none of these
Answer: (c) \( \frac{\pi}{3} \)
In simple words: Since \( \frac{2\pi}{3} \) lies outside the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) for the principal value of sine, we must first express it as \( \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) \). Then \( \sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3} \).

Exam Tip: Always check if the angle is within the principal value range. If not, use reduction formulas like \( \sin(\pi - \theta) = \sin\theta \) to bring it back into range before applying the inverse function.

Question 12. Mark the tick against the correct answer in the following: The value of \( \cos^{-1}\left(\cos\frac{13\pi}{6}\right) \) is
(a) \( \frac{13\pi}{6} \)
(b) \( \pi \)
(c) \( \frac{5\pi}{6} \)
(d) \( \frac{\pi}{6} \) and \( \frac{7\pi}{6} \)
Answer: (c) \( \frac{\pi}{6} \)
In simple words: The angle \( \frac{13\pi}{6} \) is not in the principal value range \( [0, \pi] \). We reduce it by subtracting \( 2\pi \): \( \frac{13\pi}{6} - 2\pi = \frac{\pi}{6} \). Since \( \cos^{-1}\left(\cos\frac{\pi}{6}\right) = \frac{\pi}{6} \), this is our answer.

Exam Tip: For cosine inverse, reduce angles outside \( [0, \pi] \) by adding or subtracting multiples of \( 2\pi \) until the angle falls within the principal range.

Question 13. Mark the tick against the correct answer in the following: The value of \( \tan^{-1}\left(\tan\frac{7\pi}{6}\right) \) is
(a) \( \frac{7\pi}{6} \)
(b) \( \frac{5\pi}{6} \)
(c) \( \frac{\pi}{6} \)
(d) none of these
Answer: (c) \( \frac{\pi}{6} \)
In simple words: Since \( \frac{7\pi}{6} \) is outside the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we use the property \( \tan(\pi + \theta) = \tan\theta \). So \( \tan\frac{7\pi}{6} = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\frac{\pi}{6} \). Therefore, \( \tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{\pi}{6} \).

Exam Tip: Remember that tangent has period \( \pi \), not \( 2\pi \). Use \( \tan(\pi + \theta) = \tan\theta \) or \( \tan(-\theta) = -\tan\theta \) to adjust angles into the principal range.

Question 14. Mark the tick against the correct answer in the following: The value of \( \cot^{-1}\left(\cot\frac{5\pi}{4}\right) \) is
(a) \( \frac{\pi}{4} \)
(b) \( -\frac{\pi}{4} \)
(c) \( \frac{3\pi}{4} \)
(d) none of these
Answer: (c) \( \frac{\pi}{4} \)
In simple words: The principal value range for cotangent inverse is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Since \( \frac{5\pi}{4} \) lies outside this range, we apply \( \cot(\pi + \theta) = \cot\theta \) to get \( \cot\frac{5\pi}{4} = \cot\left(\pi + \frac{\pi}{4}\right) = \cot\frac{\pi}{4} \). Thus, \( \cot^{-1}\left(\cot\frac{5\pi}{4}\right) = \frac{\pi}{4} \).

Exam Tip: Cotangent inverse has the same period property as tangent. Always reduce the angle using \( \cot(\pi + \theta) = \cot\theta \) until it falls within the principal range.

Question 15. Mark the tick against the correct answer in the following: The value of \( \sec^{-1}\left(\sec\frac{8\pi}{5}\right) \) is
(a) \( \frac{2\pi}{5} \)
(b) \( \frac{3\pi}{5} \)
(c) \( \frac{8\pi}{5} \)
(d) none of these
Answer: (a) \( \frac{2\pi}{5} \)
In simple words: The principal value range for secant inverse is \( [0, \pi] \) (excluding \( \frac{\pi}{2} \)). Since \( \frac{8\pi}{5} \) is outside this range, we reduce it by subtracting \( 2\pi \): \( \frac{8\pi}{5} - 2\pi = \frac{-2\pi}{5} \). Then we use \( \sec(2\pi - \theta) = \sec\theta \) to get \( \sec\frac{-2\pi}{5} = \sec\frac{2\pi}{5} \). Thus, the answer is \( \frac{2\pi}{5} \).

Exam Tip: For secant and cosine inverse, use multiples of \( 2\pi \) for reduction. Remember that \( \sec(2\pi - \theta) = \sec\theta \) helps convert negative or large angles.

Question 16. Mark the tick against the correct answer in the following: The value of \( \csc^{-1}\left(\csc\frac{4\pi}{3}\right) \) is
(a) \( \frac{\pi}{3} \)
(b) \( -\frac{\pi}{3} \)
(c) \( \frac{2\pi}{3} \)
(d) none of these
Answer: (b) \( -\frac{\pi}{3} \)
In simple words: The principal value range for cosecant inverse is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) (excluding 0). Since \( \frac{4\pi}{3} \) is outside this range, we use \( \csc(\pi + \theta) = -\csc\theta \) to get \( \csc\frac{4\pi}{3} = \csc\left(\pi + \frac{\pi}{3}\right) = -\csc\frac{\pi}{3} \). Therefore, \( \csc^{-1}\left(-\csc\frac{\pi}{3}\right) = -\frac{\pi}{3} \).

Exam Tip: Cosecant inverse has a range that includes negative values. Use the identity \( \csc(\pi + \theta) = -\csc\theta \) when dealing with angles in the third and fourth quadrants.

Question 17. Mark the tick against the correct answer in the following: The value of \( \tan^{-1}\left(\tan\frac{3\pi}{4}\right) \) is
(a) \( \frac{3\pi}{4} \)
(b) \( \frac{\pi}{4} \)
(c) \( -\frac{\pi}{4} \)
(d) none of these
Answer: (b) \( -\frac{\pi}{4} \)
In simple words: The angle \( \frac{3\pi}{4} \) is not in the principal range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). We apply \( \tan(\pi - \theta) = -\tan\theta \) to get \( \tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right) = -\tan\frac{\pi}{4} \). Thus, \( \tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \tan^{-1}\left(-\tan\frac{\pi}{4}\right) = -\frac{\pi}{4} \).

Exam Tip: For angles in the second quadrant, use \( \tan(\pi - \theta) = -\tan\theta \) to obtain the equivalent negative angle within the principal range.

Question 18. Mark the tick against the correct answer in the following: \( \frac{\pi}{3} - \sin^{-1}\left(\frac{-1}{2}\right) = ? \)
(a) 0
(b) \( \frac{2\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \pi \)
Answer: (b) \( \frac{\pi}{2} \)
In simple words: We know that \( \sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6} \) (using the odd function property). Substituting, \( \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi + \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \).

Exam Tip: Use the odd function property: \( \sin^{-1}(-x) = -\sin^{-1}(x) \). This simplifies calculations involving negative arguments in inverse sine.

Question 19. Mark the tick against the correct answer in the following: The value of \( \sin\left(\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}\right) = ? \)
(a) 0
(b) 1
(c) -1
(d) none of these
Answer: (b) 1
In simple words: We have \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \) and \( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \). Adding them, \( \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \). Therefore, \( \sin\left(\frac{\pi}{2}\right) = 1 \).

Exam Tip: Recall the standard values: \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \), \( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \), \( \sin\frac{\pi}{2} = 1 \). Memorizing these saves time during the exam.

Question 20. Mark the tick against the correct answer in the following: If \( x \neq 0 \) then \( \cos(\tan^{-1} x + \cot^{-1} x) = ? \)
(a) -1
(b) 1
(c) 0
(d) none of these
Answer: (c) 0
In simple words: There is a fundamental property: for any \( x \neq 0 \), \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \). Therefore, \( \cos\left(\frac{\pi}{2}\right) = 0 \).

Exam Tip: This is a key identity to memorize: \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \) for all \( x \neq 0 \). Similarly, \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).

Question 21. Mark the tick against the correct answer in the following: The value of \( \sin\left(\cos^{-1}\frac{3}{5}\right) \) is
(a) \( \frac{2}{5} \)
(b) \( \frac{4}{5} \)
(c) \( -\frac{2}{5} \)
(d) none of these
Answer: (b) \( \frac{4}{5} \)
In simple words: Let \( x = \cos^{-1}\frac{3}{5} \), so \( \cos x = \frac{3}{5} \). Using the identity \( \sin^2 x + \cos^2 x = 1 \), we get \( \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \). Since the principal value range of cosine inverse is \( [0, \pi] \) where sine is non-negative, \( \sin x = \frac{4}{5} \). Therefore, \( \sin\left(\cos^{-1}\frac{3}{5}\right) = \frac{4}{5} \).

Exam Tip: When combining inverse and forward trigonometric functions, use the Pythagorean identity and the known range of the inverse function to determine the sign of the result.

Question 22. Mark the tick against the correct answer in the following: \( \cos^{-1}\left(\cos\frac{2\pi}{3}\right) + \sin^{-1}\left(\sin\frac{2\pi}{3}\right) = ? \)
(a) \( \frac{4\pi}{3} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{5\pi}{3} \)
(d) \( \pi \)
Answer: (d) \( \pi \)
In simple words: First, \( \cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{2\pi}{3} \) since \( \frac{2\pi}{3} \) lies within \( [0, \pi] \). Second, since \( \frac{2\pi}{3} \) is outside \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we express it as \( \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \), so \( \sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3} \). Adding them: \( \frac{2\pi}{3} + \frac{\pi}{3} = \pi \).

Exam Tip: Always check the principal value range before simplifying. Use reduction formulas like \( \sin(\pi - \theta) = \sin\theta \) to bring angles into the correct range.

Question 23. Mark the tick against the correct answer in the following: \( \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = ? \)
(a) \( \frac{\pi}{3} \)
(b) \( -\frac{\pi}{3} \)
(c) \( \frac{5\pi}{3} \)
(d) none of these
Answer: (b) \( -\frac{\pi}{3} \)
In simple words: We have \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \). For \( \sec^{-1}(-2) \), we note that if \( \sec x = -2 \), then \( \cos x = -\frac{1}{2} \). The principal value of cosine inverse for \( -\frac{1}{2} \) is \( \frac{2\pi}{3} \) (since the range is \( [0, \pi] \)), so \( \sec^{-1}(-2) = \frac{2\pi}{3} \). Therefore, \( \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \).

Exam Tip: Remember that \( \sec^{-1}(y) = \cos^{-1}\left(\frac{1}{y}\right) \). For negative arguments, ensure the result stays within the principal range \( [0, \pi] \) excluding \( \frac{\pi}{2} \).

Question 24. Mark the tick against the correct answer in the following: \( \cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2} = ? \)
(a) \( \frac{2\pi}{3} \)
(b) \( \frac{3\pi}{2} \)
(c) \( 2\pi \)
(d) none of these
Answer: (a) \( \frac{2\pi}{3} \)
In simple words: We have \( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \) and \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \). So, \( \frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \).

Exam Tip: Memorize the principal values for common fractions: \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \), \( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \), \( \tan^{-1}(1) = \frac{\pi}{4} \).

Question 25. Mark the tick against the correct answer in the following: \( \tan^{-1}1 + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = ? \)
(a) \( \pi \)
(b) \( \frac{2\pi}{3} \)
(c) \( \frac{3\pi}{4} \)
(d) \( \frac{\pi}{2} \)
Answer: (c) \( \frac{3\pi}{4} \)
In simple words: We compute \( \tan^{-1}(1) = \frac{\pi}{4} \), \( \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \) (since cosine is negative in the second quadrant), and \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \) (using the odd property). Adding them: \( \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} \).

Exam Tip: Use the odd property for sine: \( \sin^{-1}(-x) = -\sin^{-1}(x) \). For cosine, directly find the angle in \( [0, \pi] \) where the cosine value matches.

Question 26. Mark the tick against the correct answer in the following: \( \tan\left(2\tan^{-1}\frac{1}{5} - \frac{\pi}{4}\right) = ? \)
(a) \( \frac{7}{17} \)
(b) \( -\frac{7}{17} \)
(c) \( \frac{7}{12} \)
(d) \( -\frac{7}{12} \)
Answer: (a) \( -\frac{7}{17} \)
In simple words: Using the double angle formula, \( \tan(2\tan^{-1}\frac{1}{5}) = \frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} = \frac{\frac{2}{5}}{\frac{24}{25}} = \frac{2}{5} \cdot \frac{25}{24} = \frac{5}{12} \). Next, apply the difference formula: \( \tan\left(\frac{5}{12} - \frac{\pi}{4}\right) = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \cdot 1} = \frac{\frac{5 - 12}{12}}{\frac{12 + 5}{12}} = \frac{-7}{17} = -\frac{7}{17} \).

Exam Tip: Apply the double angle formula \( \tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} \) first, then use the tangent difference formula carefully to combine angles.

Question 27. Mark the tick against the correct answer in the following: \( \tan\left(\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right) = ? \)
(a) \( \frac{3 - \sqrt{5}}{2} \)
(b) \( \frac{3 + \sqrt{5}}{2} \)
(c) \( \frac{5 - \sqrt{3}}{2} \)
(d) \( \frac{5 + \sqrt{3}}{2} \)
Answer: (a) \( \frac{3 - \sqrt{5}}{2} \)
In simple words: Let \( x = \cos^{-1}\frac{\sqrt{5}}{3} \), so \( \cos x = \frac{\sqrt{5}}{3} \). We use the half-angle formula: \( \tan\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}} \). Rationalizing by multiplying by \( \frac{3 - \sqrt{5}}{3 - \sqrt{5}} \), we obtain \( \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}} = \sqrt{\frac{(3 - \sqrt{5})^2}{4}} = \frac{3 - \sqrt{5}}{2} \).

Exam Tip: The half-angle formula \( \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \) is essential for this type of problem. Rationalize the denominator to simplify the final expression.

Question 28. Mark the tick against the correct answer in the following: \( \sin\left(\cos^{-1}\frac{3}{5}\right) = ? \)
(a) \( \frac{3}{4} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{3}{5} \)
(d) none of these
Answer: (b) \( \frac{4}{5} \)
In simple words: Let \( x = \cos^{-1}\frac{3}{5} \), so \( \cos x = \frac{3}{5} \). Using \( \sin^2 x + \cos^2 x = 1 \), we get \( \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} \). Since \( x \in [0, \pi] \) (the range of cosine inverse) and sine is non-negative in this range, \( \sin x = \frac{4}{5} \).

Exam Tip: Always recall the Pythagorean identity and the sign convention within the principal value range of the inverse function being used.

Question 29. Mark the tick against the correct answer in the following: \( \cos\left(\tan^{-1}\frac{3}{4}\right) = ? \)
(a) \( \frac{3}{5} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{4}{9} \)
(d) none of these
Answer: (b) \( \frac{4}{5} \)
In simple words: Let \( x = \tan^{-1}\frac{3}{4} \), so \( \tan x = \frac{3}{4} \), meaning the opposite side is 3 and the adjacent side is 4. By the Pythagorean theorem, the hypotenuse is \( \sqrt{3^2 + 4^2} = 5 \). Therefore, \( \cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \).

Exam Tip: For tangent inverse, construct a right triangle with opposite and adjacent sides matching the given ratio, then use the Pythagorean theorem to find the hypotenuse and compute the desired trigonometric ratio.

Question 30. Mark the tick against the correct answer in the following: \( \sin\left(\frac{\pi}{3} - \sin^{-1}\left(\frac{-1}{2}\right)\right) = ? \)
(a) 1
(b) 0
(c) \( -\frac{1}{2} \)
(d) none of these
Answer: (a) 1
In simple words: We have \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \) (using the odd property). So, \( \sin\left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin\left(\frac{2\pi + \pi}{6}\right) = \sin\frac{\pi}{2} = 1 \).

Exam Tip: Leverage the odd property of inverse sine to convert negative arguments into positive ones, then simplify the resulting angle sum.

Question 31. Mark the tick against the correct answer in the following: \( \sin\left(\frac{1}{2}\cos^{-1}\frac{4}{5}\right) = ? \)
(a) \( \frac{1}{\sqrt{5}} \)
(b) \( \frac{2}{\sqrt{5}} \)
(c) \( \frac{1}{\sqrt{10}} \)
(d) \( \frac{2}{\sqrt{10}} \)
Answer: (c) \( \frac{1}{\sqrt{10}} \)
In simple words: Let \( x = \cos^{-1}\frac{4}{5} \), so \( \cos x = \frac{4}{5} \). Using the half-angle formula, \( \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} \).

Exam Tip: The half-angle formula \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos\theta}{2}} \) is useful when the inverse function gives you the cosine value directly.

Question 32. Mark the tick against the correct answer in the following: \( \tan^{-1} \left\{ 2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right\} = ? \)
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{3\pi}{4} \)
(d) \( \frac{2\pi}{3} \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: Start by finding the inner inverse sine, simplify the cosine expression, and then apply the inverse tangent formula. The final result equals \( \frac{\pi}{4} \).

Exam Tip: Evaluate inverse functions from the innermost layer outward, using standard inverse trigonometric identities and Pythagoras' theorem when needed.

Question 33. Mark the tick against the correct answer in the following: If \( \cot^{-1}\left(\frac{-1}{5}\right) = x \) then \( \sin x = ? \)
(a) \( \frac{1}{\sqrt{26}} \)
(b) \( \frac{5}{\sqrt{26}} \)
(c) \( \frac{1}{\sqrt{24}} \)
(d) none of these
Answer: (b) \( \frac{5}{\sqrt{26}} \)
In simple words: When the inverse cotangent is given, treat it as an angle. Use the cotangent ratio to find the right triangle's sides, apply the Pythagorean theorem to get the hypotenuse, then calculate sine as opposite divided by hypotenuse.

Exam Tip: Always draw a reference right triangle when given an inverse trigonometric value - label the adjacent and opposite sides from the cotangent ratio, then find the hypotenuse.

Question 34. Mark the tick against the correct answer in the following: \( \sin^{-1}\left(\frac{-1}{2}\right) + 2\cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \pi \)
(c) \( \frac{3\pi}{2} \)
(d) none of these
Answer: (c) \( \frac{3\pi}{2} \)
In simple words: Evaluate the inverse sine and inverse cosine functions separately using their standard values, then combine them according to the given equation.

Exam Tip: Remember that \( \sin^{-1}(-\theta) = -\sin^{-1}(\theta) \) and \( \cos^{-1}(-\theta) = \pi - \cos^{-1}(\theta) \) when working with negative arguments.

Question 35. Mark the tick against the correct answer in the following: \( \tan^{-1}(-1) + \cos^{-1}\left(\frac{-1}{\sqrt{2}}\right) = ? \)
(a) \( \frac{\pi}{2} \)
(b) \( \pi \)
(c) \( \frac{3\pi}{2} \)
(d) \( \frac{2\pi}{3} \)
Answer: (a) \( \frac{\pi}{2} \)
In simple words: Find each inverse function value using the negative argument properties, then add them together.

Exam Tip: For negative inputs to inverse functions, apply the symmetry properties: \( \tan^{-1}(-x) = -\tan^{-1}(x) \) and \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \).

Question 36. Mark the tick against the correct answer in the following: \( \cot\left(\tan^{-1}x + \cot^{-1}x\right) = ? \)
(a) 1
(b) \( \frac{1}{2} \)
(c) 0
(d) none of these
Answer: (c) 0
In simple words: The sum \( \tan^{-1}x + \cot^{-1}x \) always equals \( \frac{\pi}{2} \), and the cotangent of \( \frac{\pi}{2} \) is zero.

Exam Tip: Memorize the complementary inverse trigonometric identity: \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \) for all real x.

Question 37. Mark the tick against the correct answer in the following: \( \tan^{-1}1 + \tan^{-1}\frac{1}{3} = ? \)
(a) \( \tan^{-1}\frac{4}{3} \)
(b) \( \tan^{-1}\frac{2}{3} \)
(c) \( \tan^{-1}2 \)
(d) \( \tan^{-1}3 \)
Answer: (a) \( \tan^{-1}\frac{4}{3} \)
In simple words: Use the addition formula for inverse tangent: when adding two inverse tangent values, apply the formula \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).

Exam Tip: This addition formula only holds when \( xy < 1 \); always check this condition before applying it.

Question 38. Mark the tick against the correct answer in the following: \( \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = ? \)
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{2\pi}{3} \)
Answer: (b) \( \frac{\pi}{4} \)
In simple words: Apply the inverse tangent addition formula to combine the two terms, then simplify to get a result that matches one of the standard angle values.

Exam Tip: When the sum formula yields an inverse tangent of 1, remember that \( \tan^{-1}(1) = \frac{\pi}{4} \).

Question 39. Mark the tick against the correct answer in the following: \( 2\tan^{-1}\frac{1}{3} = ? \)
(a) \( \tan^{-1}\frac{3}{2} \)
(b) \( \tan^{-1}\frac{3}{4} \)
(c) \( \tan^{-1}\frac{4}{3} \)
(d) none of these
Answer: (c) \( \tan^{-1}\frac{4}{3} \)
In simple words: When doubling an inverse tangent, use the double angle formula: \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \).

Exam Tip: The double angle formula for inverse tangent applies only when \( x^2 < 1 \), ensuring the result stays within the principal range.

Question 40. Mark the tick against the correct answer in the following: \( \cos\left(2\tan^{-1}\frac{1}{2}\right) = ? \)
(a) \( \frac{3}{5} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{7}{8} \)
(d) none of these
Answer: (a) \( \frac{3}{5} \)
In simple words: First find the double angle value using the addition formula for inverse tangent, then determine the cosine of that angle by constructing a right triangle from the tangent value and using the Pythagorean theorem.

Exam Tip: For expressions like \( \cos(2\tan^{-1}x) \), convert to a right triangle with tangent = x, find the hypotenuse, and directly read off the cosine ratio.

Question 41. Mark the tick against the correct answer in the following: \( \sin\left(2\tan^{-1}\frac{5}{8}\right) = ? \)
(a) \( \frac{25}{64} \)
(b) \( \frac{80}{89} \)
(c) \( \frac{75}{128} \)
(d) none of these
Answer: (b) \( \frac{80}{89} \)
In simple words: Use the double angle formula for sine: \( \sin(2\theta) = 2\sin\theta\cos\theta \). First set up a right triangle from the inverse tangent value, then apply the double angle identity.

Exam Tip: When working with \( \sin(2\tan^{-1}x) \), the formula simplifies to \( \frac{2x}{1+x^2} \), which you can derive and memorize.

Question 42. Mark the tick against the correct answer in the following: \( \sin\left(2\sin^{-1}\frac{4}{5}\right) = ? \)
(a) \( \frac{12}{25} \)
(b) \( \frac{16}{25} \)
(c) \( \frac{24}{25} \)
(d) None of these
Answer: (c) \( \frac{24}{25} \)
In simple words: For an inverse sine input, set up a right triangle where sine = \( \frac{4}{5} \), find the missing side using Pythagoras' theorem, then apply the double angle formula \( \sin(2\theta) = 2\sin\theta\cos\theta \).

Exam Tip: The key step is finding the cosine value from the right triangle (opposite and hypotenuse are given by the inverse sine), then multiplying by the sine using the double angle formula.

Question 43. Mark the tick against the correct answer in the following: If \( \tan^{-1}x = \frac{\pi}{4} - \tan^{-1}\frac{1}{3} \) then \( x = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{6} \)
(d) None of these
Answer: (a) \( \frac{1}{2} \)
In simple words: Rearrange the equation to collect both inverse tangent terms on one side, then use the subtraction formula for inverse tangent to solve for x.

Exam Tip: Use the identity \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \) and remember that \( \tan^{-1}(1) = \frac{\pi}{4} \).

Question 44. Mark the tick against the correct answer in the following: If \( \tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2} \) then \( x = ? \)
(a) 1
(b) -1
(c) 0
(d) \( \frac{1}{2} \)
Answer: (c) 0
In simple words: Apply the addition formula for inverse tangent to combine the two terms, set the result equal to \( \frac{\pi}{2} \), and solve the resulting equation algebraically.

Exam Tip: When \( \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \), this implies ab = 1 for positive arguments; use this relationship to simplify your work.

Question 45. Mark the tick against the correct answer in the following: If \( \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} \) then \( (\cos^{-1}x + \cos^{-1}y) = ? \)
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \pi \)
(d) \( \frac{2\pi}{3} \)
Answer: (b) \( \frac{\pi}{3} \)
In simple words: Use the complementary relationship between inverse sine and inverse cosine: \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \). Apply this identity separately for both x and y, then subtract the given equation from the sum of these two identities.

Exam Tip: Remember that \( \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \) and \( \cos^{-1}y = \frac{\pi}{2} - \sin^{-1}y \) for all x, y in the appropriate domains.

Question 46. Mark the tick against the correct answer in the following: \( (\tan^{-1}2 + \tan^{-1}3) = ? \)
(a) \( -\frac{\pi}{4} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{3\pi}{4} \)
(d) \( \pi \)
Answer: (c) \( \frac{3\pi}{4} \)
In simple words: Apply the addition formula for inverse tangent. Since the product of the two arguments exceeds 1, use the adjusted formula that accounts for the principal range of inverse tangent.

Exam Tip: When \( xy > 1 \) in the formula \( \tan^{-1}x + \tan^{-1}y \), add \( \pi \) to the result if both x and y are positive, placing the answer in the correct principal range.

Question 47. Mark the tick against the correct answer in the following: If \( \tan^{-1}x + \tan^{-1}3 = \tan^{-1}8 \) then \( x = ? \)
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{5} \)
(c) 3
(d) 5
Answer: (b) \( \frac{1}{5} \)
In simple words: Rearrange by moving one inverse tangent term to the right side, then apply the subtraction formula for inverse tangent to isolate and solve for x.

Exam Tip: The subtraction formula \( \tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \) is essential here; after applying it, equate the arguments on both sides.

Question 48. Mark the tick against the correct answer in the following: If \( \tan^{-1}3x + \tan^{-1}2x = \frac{\pi}{4} \) then \( x = ? \)
(a) \( \frac{1}{2} \) or -2
(b) \( \frac{1}{3} \) or -3
(c) \( \frac{1}{4} \) or -2
(d) \( \frac{1}{6} \) or -1
Answer: (a) \( \frac{1}{2} \) or -2
In simple words: Apply the addition formula for inverse tangent, set the result equal to \( \frac{\pi}{4} \), equate the tangent of both sides (using \( \tan(\pi/4) = 1 \)), and solve the resulting quadratic equation.

Exam Tip: Quadratic equations arising from inverse trigonometric expressions often yield two solutions; check both values satisfy the domain restrictions of the original equation.

Question 49. Mark the tick against the correct answer in the following: \( \tan\left(\cos^{-1}\frac{4}{5} + \tan^{-1}\frac{2}{3}\right) = ? \)
(a) \( \frac{13}{6} \)
(b) \( \frac{17}{6} \)
(c) \( \frac{19}{6} \)
(d) \( \frac{23}{6} \)
Answer: (a) \( \frac{13}{6} \)
In simple words: Convert the inverse cosine to an inverse tangent using a right triangle, apply the addition formula to combine both inverse tangent values, and simplify.

Exam Tip: To convert \( \cos^{-1}\frac{4}{5} \) to an inverse tangent, draw a right triangle, use the Pythagorean theorem to find the missing side, and determine the tangent ratio.

Question 50. Mark the tick against the correct answer in the following: \( \tan\left(\cos^{-1}\frac{4}{5} + \tan^{-1}\frac{2}{3}\right) = ? \)
(a) \( \frac{13}{6} \)
(b) \( \frac{17}{6} \)
(c) \( \frac{19}{6} \)
(d) \( \frac{23}{6} \)
Answer: (b) \( \frac{17}{6} \)
In simple words: Create a right triangle from the inverse cosine value using the cosine ratio, calculate the opposite side using the Pythagorean theorem, and convert to an inverse tangent. Then combine with the second inverse tangent using the addition formula.

Exam Tip: After converting \( \cos^{-1}\frac{4}{5} \) to \( \tan^{-1}\frac{3}{4} \), carefully apply the addition formula and simplify the resulting fractions.

Question 51. Mark the tick against the correct answer in the following: Range of \( \sin^{-1}x \) is
(a) \( \left[0, \frac{\pi}{2}\right] \)
(b) \( [0, \pi] \)
(c) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(d) None of these
Answer: (c) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
In simple words: The inverse sine function has a restricted output (range) to ensure it is a one-to-one function. This range extends from negative \( \pi/2 \) to positive \( \pi/2 \), covering both the lower and upper quadrants symmetrically around the origin.

Exam Tip: The range of \( \sin^{-1}x \) must be memorized as a fundamental property; it is symmetric about zero and corresponds to a quarter-circle on each side of the origin.

Question 52. Mark the tick against the correct answer in the following: Range of \( \cos^{-1}x \) is
(a) \( [0, \pi] \)
(b) \( \left[0, \frac{\pi}{2}\right] \)
(c) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(d) None of these
Answer: (a) \( [0, \pi] \)
In simple words: The inverse cosine function is defined to output values only in the range from zero to \( \pi \). This choice ensures the function is one-to-one and covers all possible input values between -1 and 1 without repetition.

Exam Tip: Unlike inverse sine, the range of \( \cos^{-1}x \) is entirely non-negative and spans from 0 (when x = 1) to \( \pi \) (when x = -1).

Question 53. Mark the tick against the correct answer in the following: Range of \( \tan^{-1}x \) is
(a) \( \left[0, \frac{\pi}{2}\right] \)
(b) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(c) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(d) None of these
Answer: (b) \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
In simple words: The inverse tangent function outputs values strictly between negative and positive \( \pi/2 \), never actually reaching these boundary values. The endpoints are approached but never attained as x approaches negative and positive infinity.

Exam Tip: The range of \( \tan^{-1}x \) uses open brackets (parentheses) at the endpoints because the asymptotic values \( \pm\frac{\pi}{2} \) are never actually achieved.

Question 54. Mark the tick against the correct answer in the following: Range of \( \sec^{-1}x \) is
(a) \( \left[0, \frac{\pi}{2}\right] \)
(b) \( [0, \pi] \)
(c) \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \)
(d) None of these
Answer: (c) \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \)
In simple words: The inverse secant function is defined on the full interval from 0 to \( \pi \), with one critical exception: the value \( \pi/2 \) is excluded because secant is undefined at that point (cosine equals zero there).

Exam Tip: Remember that \( \sec^{-1}x = \cos^{-1}\left(\frac{1}{x}\right) \); since cosine of \( \pi/2 \) is zero, no input would produce that output, so it must be excluded.

Question 55. Mark the tick against the correct answer in the following: Range of cosec⁻¹ x is
(a) \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
(b) \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)
(c) \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \)
(d) None of the options
Answer: (c) \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \)
In simple words: The inverse cosecant function outputs all values from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) using closed brackets, but we must exclude 0 because cosecant is undefined there.

Exam Tip: Remember that the range excludes 0 because the original cosecant function never equals zero - this restriction carries over to its inverse.

Question 56. Mark the tick against the correct answer in the following: Domain of cos⁻¹ x is
(a) [0, 1]
(b) [-1, 1]
(c) [-1, 0]
(d) None of the options
Answer: (b) [-1, 1]
In simple words: The inverse cosine function accepts any input value between -1 and 1, inclusive. No values outside this range can be entered into the function.

Exam Tip: The domain of inverse cosine mirrors the range of the original cosine function, which is always bounded between -1 and 1.

Question 57. Mark the tick against the correct answer in the following: Domain of sec⁻¹ x is
(a) [-1, 1]
(b) R - {0}
(c) R - [-1, 1]
(d) R - {-1, 1}
Answer: (c) R - [-1, 1]
In simple words: The inverse secant function accepts any real number except those between -1 and 1 inclusive. All values greater than 1 or less than -1 are valid inputs.

Exam Tip: Think of it as the opposite of inverse cosine - instead of accepting values only between -1 and 1, inverse secant rejects that entire interval.