CBSE Class 12 Mathematics Inverse Trigonometric Functions VBQs Set 03

Question. Fill in the blanks.
(i) The principal value of \( \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \) is _____________ .
(ii) The set of values of \( \sec^{-1} \left( \frac{1}{2} \right) \) is _____________ .
(iii) The value of \( \cos(\sin^{-1} x + \cos^{-1} x), |x| \le 1 \) is _____________ .
(iv) The value of \( \sin^{-1} \left( \cos \left( \frac{43\pi}{5} \right) \right) \) is _____________ .
Answer: (i) \( \frac{5\pi}{6} \), (ii) \( \phi \), (iii) \( 0 \), (iv) \( -\frac{\pi}{10} \)

Very Short Answer Questions:

Question. What is the domain of the function \( \sin^{-1} x \)?
Answer: \( [-1, 1] \)

Question. Write the principal value of \( \cot^{-1}(-\sqrt{3}) \).
Answer: \( \frac{5\pi}{6} \)

Question. If \( 4 \cos^{-1} x + \sin^{-1} x = \pi \), then find the value of \( x \).
Answer: \( x = \frac{\sqrt{3}}{2} \)

Question. Evaluate: \( \tan (\tan^{-1} (-4)) \)
Answer: \( -4 \)

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

Question. Write the value of \( \sin(\cot^{-1} x) \).
Answer: \( \frac{1}{\sqrt{1 + x^2}} \)

Short Answer Questions-I:

Question. Find the value of \( \sin(2 \sin^{-1} (0.6)) \).
Answer: \( 0.96 \)

Question. Show that \( \sin^{-1}(2x\sqrt{1-x^2}) = 2 \sin^{-1} x, -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \).
Answer: Let \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
LHS \( = \sin^{-1}(2\sin \theta \sqrt{1 - \sin^2 \theta}) \)
\( = \sin^{-1}(2\sin \theta \cos \theta) \)
\( = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \sin^{-1} x \).
Hence Proved.

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

Question. Prove that: \( 3 \sin^{-1} x = \sin^{-1}(3x - 4x^3), x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] \).
Answer: Let \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
RHS \( = \sin^{-1}(3\sin \theta - 4\sin^3 \theta) \)
\( = \sin^{-1}(\sin 3\theta) = 3\theta = 3 \sin^{-1} x \).
Hence Proved.

Question. Write the simplest form of \( \tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right), |x| < a \).
Answer: \( \sin^{-1} \frac{x}{a} \)

Question. Write the principal value of \( \tan^{-1} \sqrt{3} - \cot^{-1}(-\sqrt{3}) \).
Answer: \( -\frac{\pi}{2} \)

Short Answer Questions-II:

Question. Prove that: \( \cos \left( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \right) = \frac{6}{5\sqrt{13}} \).
Answer: Let \( \sin^{-1} \frac{3}{5} = A \Rightarrow \sin A = \frac{3}{5}, \cos A = \frac{4}{5} \).
Let \( \cot^{-1} \frac{3}{2} = B \Rightarrow \cot B = \frac{3}{2} \Rightarrow \sin B = \frac{2}{\sqrt{13}}, \cos B = \frac{3}{\sqrt{13}} \).
\( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
\( = \frac{4}{5} \cdot \frac{3}{\sqrt{13}} - \frac{3}{5} \cdot \frac{2}{\sqrt{13}} = \frac{12 - 6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}} \).
Hence Proved.

Question. Solve: \( \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1} \frac{8}{31} \).
Answer: \( x = \frac{1}{4} \)

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

Question. Evaluate: \( \tan \left\{ 2 \tan^{-1} \left( \frac{1}{5} \right) + \frac{\pi}{4} \right\} \).
Answer: \( \frac{17}{7} \)

Question. Prove the following:
\( \cot^{-1} \left( \frac{xy + 1}{x - y} \right) + \cot^{-1} \left( \frac{yz + 1}{y - z} \right) + \cot^{-1} \left( \frac{zx + 1}{z - x} \right) = 0 \) (\( 0 < xy, yz, zx < 1 \)).
Answer: RHS \( = (\tan^{-1} x - \tan^{-1} y) + (\tan^{-1} y - \tan^{-1} z) + (\tan^{-1} z - \tan^{-1} x) = 0 \).
Hence Proved.

Question. Prove the following:
\( \sin \left[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right] = 1, 0 < x < 1 \).
Answer: Let \( x = \tan \theta \).
\( = \sin[\tan^{-1}(\cot 2\theta) + \cos^{-1}(\cos 2\theta)] \)
\( = \sin[\tan^{-1}(\tan(\frac{\pi}{2} - 2\theta)) + 2\theta] \)
\( = \sin[\frac{\pi}{2} - 2\theta + 2\theta] = \sin \frac{\pi}{2} = 1 \).
Hence Proved.

Question. Prove that: \( 2 \tan^{-1} \left( \sqrt{\frac{a - b}{a + b}} \tan \frac{x}{2} \right) = \cos^{-1} \left( \frac{a \cos x + b}{a + b \cos x} \right) \).
Answer: Standard transformation of inverse trigonometric functions. Hence Proved.

Question. If \( \tan^{-1} \left( \frac{1}{1 + 1.2} \right) + \tan^{-1} \left( \frac{1}{1 + 2.3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) = \tan^{-1} \theta \), then find the value of \( \theta \).
Answer: \( \theta = \frac{n}{n + 2} \)

Question. Prove that: \( \tan^{-1} \left( \frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}} \right) = \frac{\pi}{4} + \frac{x}{2} \), where \( \pi < x < \frac{3\pi}{2} \).
Answer: By substituting \( \cos x = 2 \cos^2(x/2) - 1 \) and \( \cos x = 1 - 2 \sin^2(x/2) \). Hence Proved.

Question. Solve for \( x \): \( \tan^{-1} \left( \frac{2 - x}{2 + x} \right) = \frac{1}{2} \tan^{-1} \frac{x}{2}, x > 0 \).
Answer: \( x = \frac{2}{\sqrt{3}} \)

Question. Prove that: \( 2 \sin^{-1} \left( \frac{3}{5} \right) - \tan^{-1} \left( \frac{17}{31} \right) = \frac{\pi}{4} \).
Answer: Using \( 2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7} \) and then \( \tan^{-1} \frac{24}{7} - \tan^{-1} \frac{17}{31} = \frac{\pi}{4} \). Hence Proved.

Question. Prove 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 \( 2x = \tan \theta \). Then use the formula for \( \tan 3\theta \) and \( \tan 2\theta \). Hence Proved.

Question. Solve for \( x \): \( \tan^{-1} \left[ \frac{x - 3}{x - 4} \right] + \tan^{-1} \left[ \frac{x + 3}{x + 4} \right] = \frac{\pi}{4} \).
Answer: \( x = \pm \sqrt{\frac{17}{2}} \)

Question. If \( \tan^{-1} x - \cot^{-1} x = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right), x > 0 \), find the value of \( x \) and hence find the value of \( \sec^{-1} \left( \frac{2}{x} \right) \).
Answer: \( x = \sqrt{3} \); \( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6} \)

Question. If \( \sin^{-1} \left( \frac{3}{x} \right) + \sin^{-1} \left( \frac{4}{x} \right) = \frac{\pi}{2} \), then find the value of \( x \).
Answer: \( x = 5 \)

Question. Find the value of \( x \), if \( \tan[\sec^{-1}(1/x)] = \sin(\tan^{-1} 2), x > 0 \).
Answer: \( x = \frac{\sqrt{5}}{3} \)

Question. Prove that \( \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \sin^{-1} \left( \frac{4}{5} \right) \).
Answer: \( \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \tan^{-1} \left( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{2}{36}} \right) = \tan^{-1} \frac{1}{2} = \frac{1}{2} (2 \tan^{-1} \frac{1}{2}) = \frac{1}{2} \sin^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 + \frac{1}{4}} \right) = \frac{1}{2} \sin^{-1} \frac{4}{5} \). Hence Proved.