CBSE Class 12 Mathematics Inverse Trigonometric Functions Worksheet Set 05

Question. The domain of function defined by \( f(x) = \text{cosec}^{-1}(x) \) is
(a) \( [-1, 1] \)
(b) \( \mathbb{R} \)
(c) \( \mathbb{R} - (-1, 1) \)
(d) None of the options
Answer: (c) \( \mathbb{R} - (-1, 1) \)

Question. The range of function defined by \( f(x) = \text{cot}^{-1}x \) is
(a) \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
(b) \( [0, \pi] \)
(c) \( (0, \pi) \)
(d) \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \)
Answer: (c) \( (0, \pi) \)

Question. If \( \tan^{-1} x = y \), then
(a) \( -1 < y < 1 \)
(b) \( -\frac{\pi}{2} \le y \le \frac{\pi}{2} \)
(c) \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
(d) \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Answer: (c) \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)

Question. The domain of \( \cos^{-1}(x^2 - 4) \) is
(a) \( [-1, 1] \)
(b) \( [-2, 2] \)
(c) \( [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}] \)
(d) \( [0, \pi] \)
Answer: (c) \( [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}] \)

Question. The principal value of \( \sin^{-1}\left(\sin \frac{13\pi}{7}\right) \) is
(a) \( \frac{\pi}{7} \)
(b) \( -\frac{\pi}{7} \)
(c) \( \frac{\pi}{7} \)
(d) \( \frac{\pi}{3} \)
Answer: (b) \( -\frac{\pi}{7} \)

Question. \( \sin^{-1}\left(\cos \frac{\pi}{6}\right) \) is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{6} \)
Answer: (c) \( \frac{\pi}{3} \)

Question. The value of \( \cot(\sin^{-1} x) \) is
(a) \( \frac{\sqrt{1+x^2}}{x} \)
(b) \( \frac{x}{\sqrt{1+x^2}} \)
(c) \( \frac{1}{x} \)
(d) \( \frac{\sqrt{1-x^2}}{x} \)
Answer: (d) \( \frac{\sqrt{1-x^2}}{x} \)

Question. The value of \( \cot\left[\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right] \) is
(a) \( \frac{1}{\sqrt{3}} \)
(b) \( 1 \)
(c) \( \sqrt{3} \)
(d) \( 0 \)
Answer: (c) \( \sqrt{3} \)

Question. \( \sin\left[\frac{\pi}{3} + \sin^{-1}\left(\frac{1}{2}\right)\right] \) is equal to
(a) \( 1 \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{4} \)
Answer: (a) \( 1 \)

Question. \( \sin\left[\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right] \) is equal to
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( -1 \)
(d) \( 1 \)
Answer: (d) \( 1 \)

Assertion-Reason Based Questions

Question. Assertion (A) The domain of the function \( \text{sec}^{-1} 2x \) is \( (-\infty, -1/2] \cup [1/2, \infty) \).
Reason (R) \( \text{sec}^{-1}(-2) = -\frac{\pi}{4} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) The value of \( \sin^{-1}\left(\cos \frac{33\pi}{5}\right) \) is \( -\frac{\pi}{10} \).
Reason (R) The principal value branch of \( \sin^{-1} x \) is \( [0, \pi] \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) The value of \( \sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right] \) is \( 1 \).
Reason (R) \( \tan^{-1}(-x) = \tan^{-1} x \) and \( \cos^{-1}(-x) = \cos^{-1} x \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) \( \sin^{-1}\left(\frac{1}{2}\right) + \cot^{-1}\sqrt{3} = \frac{\pi}{3} \).
Reason (R) Principal value branch of \( \cos^{-1} x \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) \( \sin(\tan^{-1} x) \), \( |x| < 1 \) is equal to \( \frac{x}{\sqrt{1+x^2}} \).
Reason (R) Domain of \( \tan^{-1} x \) is \( (0, \pi) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Case Based Questions

Two men on either side of a temple of \( 30\text{ m} \) high observe its top at the angles of elevation \( \alpha \) and \( \beta \), respectively. The distance between the two men is \( 40\sqrt{3}\text{ m} \) and the distance between the first person A and the temple is \( 30\sqrt{3}\text{ m} \).

Question. \( \angle CAB = \alpha \) is
(a) \( \sin^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
(b) \( \sin^{-1}\left(\frac{1}{2}\right) \)
(c) \( \sin^{-1}(2) \)
(d) \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
Answer: (b) \( \sin^{-1}\left(\frac{1}{2}\right) \)

Question. \( \angle CAB = \alpha \) is
(a) \( \cos^{-1}\left(\frac{1}{5}\right) \)
(b) \( \cos^{-1}\left(\frac{2}{5}\right) \)
(c) \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
(d) \( \cos^{-1}\left(\frac{4}{5}\right) \)
Answer: (c) \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)

Question. \( \angle BCA = \beta \) is
(a) \( \tan^{-1}\left(\frac{1}{2}\right) \)
(b) \( \tan^{-1}(2) \)
(c) \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
(d) \( \tan^{-1}(\sqrt{3}) \)
Answer: (d) \( \tan^{-1}(\sqrt{3}) \)

Question. \( \angle ABC \) is
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{3} \)
Answer: (c) \( \frac{\pi}{2} \)

Question. Domain and range of \( \cos^{-1} x \) is equal to
(a) \( (-1, 1), (0, \pi) \)
(b) \( [-1, 1], (0, \pi) \)
(c) \( [-1, 1], [0, \pi] \)
(d) \( (-1, 1), \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Answer: (c) \( [-1, 1], [0, \pi] \)

Very Short Answer Type Questions

Question. Evaluate \( \cos\left[\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right] \).
Answer: \( -1 \)

Question. Find the value of \( x \), if \( \cos(2\sin^{-1} x) = \frac{1}{9} \), \( x > 0 \).
Answer: \( \frac{2}{3} \)

Question. Write the simplest form of \( \cot^{-1}\left(\frac{1}{\sqrt{x^2 - 1}}\right) \), \( |x| > 1 \).
Answer: \( \text{sec}^{-1}x \)

Question. Prove that \( 3\sin^{-1} x = \sin^{-1}(3x - 4x^3) \), \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \).
Answer: Let \( x = \sin\theta \), then \( \theta = \sin^{-1} x \).
\( \text{RHS} = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x = \text{LHS} \).
Since \( x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \), \( 3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), which is the principal value branch of \( \sin^{-1} x \). Hence proved.

Question. Write the simplest form of \( \tan^{-1}\sqrt{\frac{1 - \cos x}{1 + \cos x}} \), \( x < \pi \).
Answer: \( \frac{x}{2} \)

Short Answer Type Questions

Question. If \( \sin\left[\text{cot}^{-1}(x + 1)\right] = \cos(\tan^{-1} x) \), then find the value of \( x \).
Answer: \( x = -\frac{1}{2} \)

Question. Find the value of \( x \), if \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1} x \), \( x > 0 \).
Answer: \( x = \frac{1}{\sqrt{3}} \)

Question. Evaluate \( \cos\left[\sin^{-1}\frac{1}{4} + \text{sec}^{-1}\frac{4}{3}\right] \).
Answer: \( \frac{3\sqrt{15} - \sqrt{7}}{16} \)

Question. Show that \( \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) \), \( x \in [0, 1] \).
Answer: Let \( x = \tan^2\theta \implies \sqrt{x} = \tan\theta \implies \theta = \tan^{-1}\sqrt{x} \).
\( \text{RHS} = \frac{1}{2}\cos^{-1}\left(\frac{1 - \tan^2\theta}{1 + \tan^2\theta}\right) = \frac{1}{2}\cos^{-1}(\cos 2\theta) = \theta = \tan^{-1}\sqrt{x} = \text{LHS} \). Hence proved.

Question. Express \( \tan^{-1}\left(\frac{x}{a + \sqrt{a^2 - x^2}}\right) \) in simplest form.
Answer: \( \frac{1}{2}\sin^{-1}\left(\frac{x}{a}\right) \)

Question. Express \( \cos^{-1}\left(\frac{3}{5}\cos x + \frac{4}{5}\sin x\right) \) in simplest form.
Answer: \( x - \tan^{-1}\left(\frac{4}{3}\right) \)

Long Answer Type Questions

Question. Prove that \( \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \cos^{-1}\left(\frac{36}{85}\right) \).
Answer: Let \( \sin^{-1}\left(\frac{8}{17}\right) = x \implies \sin x = \frac{8}{17} \), \( \cos x = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \frac{15}{17} \).
Let \( \sin^{-1}\left(\frac{3}{5}\right) = y \implies \sin y = \frac{3}{5} \), \( \cos y = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).
Now, \( \cos(x + y) = \cos x\cos y - \sin x\sin y = \left(\frac{15}{17}\right)\left(\frac{4}{5}\right) - \left(\frac{8}{17}\right)\left(\frac{3}{5}\right) = \frac{60 - 24}{85} = \frac{36}{85} \).
\( \implies x + y = \cos^{-1}\left(\frac{36}{85}\right) \).
Substituting \( x \) and \( y \): \( \sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \cos^{-1}\left(\frac{36}{85}\right) \). Hence proved.

Question. Solve the following: \( 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) - 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \).
Answer: Let \( x = \tan\theta \).
Using standard inverse trigonometric identities:
\( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\theta \), \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\theta \), and \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\theta \).
Substituting these values in the equation:
\( 3(2\theta) - 4(2\theta) + 2(2\theta) = \frac{\pi}{3} \)
\( 6\theta - 8\theta + 4\theta = \frac{\pi}{3} \)
\( 2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6} \).
Since \( x = \tan\theta \), we get:
\( x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \).

Question. Prove that \( \tan^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right) = \frac{\pi}{4} + \frac{1}{2}\cos^{-1} x^2 \).
Answer: Let \( x^2 = \cos 2\theta \implies 2\theta = \cos^{-1} x^2 \implies \theta = \frac{1}{2}\cos^{-1} x^2 \).
Substitute \( x^2 \) in LHS:
\( \text{LHS} = \tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}\right) \)
Using \( 1+\cos 2\theta = 2\cos^2\theta \) and \( 1-\cos 2\theta = 2\sin^2\theta \):
\( = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\frac{1+\tan\theta}{1-\tan\theta}\right) = \tan^{-1}\left[\tan\left(\frac{\pi}{4} + \theta\right)\right] = \frac{\pi}{4} + \theta = \frac{\pi}{4} + \frac{1}{2}\cos^{-1} x^2 = \text{RHS} \). Hence proved.