CBSE Class 12 Mathematics Inverse Trigonometric Functions Worksheet Set 04

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 range of the function \( f(x) = 2\sin^{-1} x + \frac{3\pi}{2} \), where \( x \in [-1, 1] \) is \( \left[\frac{\pi}{2}, \frac{5\pi}{2}\right] \).
Reason (R) The range of 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) Maximum value of \( (\cos^{-1} x)^2 \) is \( \pi^2 \).
Reason (R) The 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) All trigonometric functions have their inverses over their respective domains.
Reason (R) The inverse of \( \tan^{-1} x \) exists for some \( x \in \mathbb{R} \).
(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: (d) R is correct; A is incorrect.

Question. Assertion (A) Domain of \( y = \cos^{-1}(x) \) is \( [-1, 1] \).
Reason (R) The range of the principal value branch of \( y = \cos^{-1}(x) \) is \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \).
(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) Range of \( [2\cos^{-1} x] \) is \( [0, \pi] \).
Reason (R) Principal value branch of \( \sin^{-1} x \) has range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
(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: (d) R is correct; A 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. Write the domain and range (principal value branch) of the following function \( f(x) = \text{sec}^{-1}x \).
Answer: Domain: \( \mathbb{R} - (-1, 1) \), Range: \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \).

Question. Find the principal value of \( \cos^{-1}\left(-\frac{1}{2}\right) \).
Answer: \( \frac{2\pi}{3} \)

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

Question. Find the value of \( \cot\left[\cos^{-1}\left(\frac{7}{25}\right)\right] \).
Answer: \( \frac{7}{24} \)

Short Answer Type Questions

Question. Find the domain of the following functions:
(i) \( \sin^{-1} 2x \)
(ii) \( \sin^{-1} x + \cos x \)
Answer:
(i) \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)
(ii) \( [-1, 1] \)

Question. If \( a = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right) \) and \( b = \tan^{-1}(-\sqrt{3}) - \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \), then find the value of \( a + b \).
Answer:
\( a = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \left(\pi - \frac{\pi}{3}\right) = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{11\pi}{12} \)
\( b = \tan^{-1}\sqrt{3} - \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} - \left(\pi - \frac{\pi}{3}\right) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \)
\( a + b = \frac{11\pi}{12} - \frac{\pi}{3} = \frac{7\pi}{12} \)

Question. Evaluate \( \sin^{-1}\left(\sin \frac{13\pi}{6}\right) + \cos^{-1}\left(\cos \frac{\pi}{3}\right) + \tan^{-1}\sqrt{3} \).
Answer:
\( = \sin^{-1}\left(\sin\left(2\pi + \frac{\pi}{6}\right)\right) + \frac{\pi}{3} + \frac{\pi}{3} \)
\( = \frac{\pi}{6} + \frac{\pi}{3} + \frac{\pi}{3} = \frac{5\pi}{6} \)

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

Question. Find the simplest form of \( \sin^{-1}(2x\sqrt{1 - x^2}) \), \( \frac{1}{\sqrt{2}} \le x \le 1 \).
Answer: \( 2\cos^{-1} x \)

Long Answer Type Questions

Question. If \( \cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \theta \), then show that \( \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2} = \sin^2\theta \).
Answer: Given \( \cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \theta \implies \cos^{-1}\frac{x}{a} = \theta - \cos^{-1}\frac{y}{b} \).
Taking cosine on both sides:
\( \frac{x}{a} = \cos\left(\theta - \cos^{-1}\frac{y}{b}\right) = \cos\theta \cos\left(\cos^{-1}\frac{y}{b}\right) + \sin\theta \sin\left(\cos^{-1}\frac{y}{b}\right) \).
Let \( \cos^{-1}\frac{y}{b} = z \implies \cos z = \frac{y}{b} \implies \sin z = \sqrt{1 - \frac{y^2}{b^2}} \).
So, \( \frac{x}{a} = \frac{y}{b}\cos\theta + \sin\theta \sqrt{1 - \frac{y^2}{b^2}} \implies \frac{x}{a} - \frac{y}{b}\cos\theta = \sin\theta \sqrt{1 - \frac{y^2}{b^2}} \).
Squaring both sides:
\( \left(\frac{x}{a} - \frac{y}{b}\cos\theta\right)^2 = \sin^2\theta \left(1 - \frac{y^2}{b^2}\right) \)
\( \frac{x^2}{a^2} + \frac{y^2}{b^2}\cos^2\theta - \frac{2xy}{ab}\cos\theta = \sin^2\theta - \frac{y^2}{b^2}\sin^2\theta \)
\( \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2}(\cos^2\theta + \sin^2\theta) = \sin^2\theta \).
Since \( \cos^2\theta + \sin^2\theta = 1 \), we get:
\( \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2} = \sin^2\theta \). Hence proved.

Question. Show that \( \tan^{-1}\left(\frac{6x - 8x^3}{1 - 12x^2}\right) - \tan^{-1}\left(\frac{4x}{1 - 4x^2}\right) = \tan^{-1} 2x \), \( |2x| < \frac{1}{\sqrt{3}} \).
Answer: Let \( 2x = \tan\theta \).
The expression can be written as:
\( \tan^{-1}\left(\frac{3(2x) - (2x)^3}{1 - 3(2x)^2}\right) - \tan^{-1}\left(\frac{2(2x)}{1 - (2x)^2}\right) \)
\( = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right) - \tan^{-1}\left(\frac{2\tan\theta}{1 - \tan^2\theta}\right) \)
\( = \tan^{-1}(\tan 3\theta) - \tan^{-1}(\tan 2\theta) = 3\theta - 2\theta = \theta = \tan^{-1} 2x \). Hence proved.