CBSE Class 12 Mathematics Inverse Trigonometric Functions VBQs Set 02

Question. Prove that if \( \frac{1}{2} \le x \le 1 \) then \( \cos^{-1} x + \cos^{-1} \left[ \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} \right] = \frac{\pi}{3} \).
Answer: Let \( \cos^{-1} x = \theta \Rightarrow x = \cos \theta \).
Now, \( \cos^{-1} x + \cos^{-1} \left[ \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} \right] = \theta + \cos^{-1} \left[ \frac{1}{2} \cdot \cos \theta + \frac{\sqrt{3}}{2} \sqrt{1 - \cos^2 \theta} \right] \)
\( = \theta + \cos^{-1} \left[ \cos \frac{\pi}{3} \cdot \cos \theta + \sin \frac{\pi}{3} \cdot \sin \theta \right] \)
\( = \theta + \cos^{-1} \left\{ \cos \left( \frac{\pi}{3} - \theta \right) \right\} \)
Since \( \frac{1}{2} \le x \le 1 \Rightarrow \cos \frac{\pi}{3} \le \cos \theta \le \cos 0 \Rightarrow 0 \le \theta \le \frac{\pi}{3} \)
\( \Rightarrow -\frac{\pi}{3} \le -\theta \le 0 \Rightarrow 0 \le \frac{\pi}{3} - \theta \le \frac{\pi}{3} \)
\( \therefore \theta + \frac{\pi}{3} - \theta = \frac{\pi}{3} \).

Question. If \( \tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}, xy < 1 \), then write the value of \( x + y + xy \).
Answer: Given \( \tan^{-1} x + \tan^{-1} y = \frac{\pi}{4} \Rightarrow \tan^{-1} \left[ \frac{x + y}{1 - xy} \right] = \frac{\pi}{4} \)
\( \Rightarrow \frac{x + y}{1 - xy} = \tan \frac{\pi}{4} = 1 \)
\( \Rightarrow x + y = 1 - xy \)
\( \Rightarrow x + y + xy = 1 \).

Short Answer Questions

Question. Show that: \( \tan \left( \frac{1}{2} \sin^{-1} \frac{3}{4} \right) = \frac{4 - \sqrt{7}}{3} \).
Answer: Let \( \sin^{-1} \frac{3}{4} = \theta \Rightarrow \sin \theta = \frac{3}{4} \) where \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
Using \( \sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \)
\( \Rightarrow \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} = \frac{3}{4} \)
\( \Rightarrow 8 \tan \frac{\theta}{2} = 3 + 3 \tan^2 \frac{\theta}{2} \Rightarrow 3 \tan^2 \frac{\theta}{2} - 8 \tan \frac{\theta}{2} + 3 = 0 \)
Using quadratic formula for \( \tan \frac{\theta}{2} \):
\( \tan \frac{\theta}{2} = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3} \)
As \( \sin \theta = \frac{3}{4} \), \( \theta \) is in the first quadrant, so \( \tan \frac{\theta}{2} = \frac{4 - \sqrt{7}}{3} \).
\( \therefore \tan \left( \frac{1}{2} \sin^{-1} \frac{3}{4} \right) = \frac{4 - \sqrt{7}}{3} \).

Question. Evaluate: \( \tan \left\{ 2 \tan^{-1} \left( \frac{1}{5} \right) + \frac{\pi}{4} \right\} \).
Answer: \( \tan \left\{ 2 \tan^{-1} \left( \frac{1}{5} \right) + \frac{\pi}{4} \right\} = \tan \left\{ \tan^{-1} \left[ \frac{2 \times \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} \right] + \tan^{-1} 1 \right\} \)
\( = \tan \left\{ \tan^{-1} \left( \frac{2/5}{24/25} \right) + \tan^{-1} 1 \right\} = \tan \left\{ \tan^{-1} \frac{5}{12} + \tan^{-1} 1 \right\} \)
\( = \tan \left\{ \tan^{-1} \left[ \frac{\frac{5}{12} + 1}{1 - \frac{5}{12} \times 1} \right] \right\} = \frac{17/12}{7/12} = \frac{17}{7} \).

Question. Prove that: \( \cot^{-1} 7 + \cot^{-1} 8 + \cot^{-1} 18 = \cot^{-1} 3 \).
Answer: LHS \( = \cot^{-1} 7 + \cot^{-1} 8 + \cot^{-1} 18 \)
\( = \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18} \)
\( = \tan^{-1} \left[ \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}} \right] + \tan^{-1} \frac{1}{18} \)
\( = \tan^{-1} \frac{15}{55} + \tan^{-1} \frac{1}{18} = \tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18} \)
\( = \tan^{-1} \left[ \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \cdot \frac{1}{18}} \right] = \tan^{-1} \left[ \frac{\frac{54 + 11}{198}}{\frac{198 - 3}{198}} \right] = \tan^{-1} \frac{65}{195} \)
\( = \tan^{-1} \frac{1}{3} = \cot^{-1} 3 = \text{RHS} \).

Question. Prove that: \( \sin^{-1} \left( \frac{63}{65} \right) = \sin^{-1} \left( \frac{5}{13} \right) + \cos^{-1} \left( \frac{3}{5} \right) \).
Answer: Let \( \sin^{-1} \frac{5}{13} = \alpha \Rightarrow \sin \alpha = \frac{5}{13}, \cos \alpha = \frac{12}{13} \)
Let \( \cos^{-1} \frac{3}{5} = \beta \Rightarrow \cos \beta = \frac{3}{5}, \sin \beta = \frac{4}{5} \)
Now, \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\( = \frac{5}{13} \cdot \frac{3}{5} + \frac{12}{13} \cdot \frac{4}{5} = \frac{15}{65} + \frac{48}{65} = \frac{63}{65} \)
\( \Rightarrow \alpha + \beta = \sin^{-1} \frac{63}{65} \)
\( \Rightarrow \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5} = \sin^{-1} \frac{63}{65} \).

Question. Prove the following: \( \cos[\tan^{-1}\{\sin(\cot^{-1} x)\}] = \sqrt{\frac{1 + x^2}{2 + x^2}} \).
Answer: Let \( \cot^{-1} x = \theta \Rightarrow x = \cot \theta \Rightarrow \sin \theta = \frac{1}{\csc \theta} = \frac{1}{\sqrt{1 + \cot^2 \theta}} = \frac{1}{\sqrt{1 + x^2}} \)
The expression becomes \( \cos[\tan^{-1} \{ \frac{1}{\sqrt{1 + x^2}} \} ] \)
Let \( \tan^{-1} \frac{1}{\sqrt{1 + x^2}} = \phi \Rightarrow \tan \phi = \frac{1}{\sqrt{1 + x^2}} \)
We know \( \cos \phi = \frac{1}{\sec \phi} = \frac{1}{\sqrt{1 + \tan^2 \phi}} \)
\( = \frac{1}{\sqrt{1 + \left( \frac{1}{\sqrt{1 + x^2}} \right)^2}} = \frac{1}{\sqrt{1 + \frac{1}{1 + x^2}}} = \frac{1}{\sqrt{\frac{1 + x^2 + 1}{1 + x^2}}} = \sqrt{\frac{1 + x^2}{2 + x^2}} = \text{RHS} \).

Question. Solve for \( x \): \( \sin^{-1}(1 - x) - 2 \sin^{-1} x = \frac{\pi}{2} \).
Answer: \( \sin^{-1}(1 - x) = \frac{\pi}{2} + 2 \sin^{-1} x \)
\( \Rightarrow (1 - x) = \sin \left( \frac{\pi}{2} + 2 \sin^{-1} x \right) = \cos(2 \sin^{-1} x) \)
\( \Rightarrow 1 - x = 1 - 2 \sin^2 (\sin^{-1} x) \)
\( \Rightarrow 1 - x = 1 - 2x^2 \Rightarrow 2x^2 - x = 0 \)
\( \Rightarrow x(2x - 1) = 0 \Rightarrow x = 0 \) or \( x = \frac{1}{2} \)
But \( x = \frac{1}{2} \) does not satisfy the given equation.
\( \therefore x = 0 \) is the required solution.

Question. Prove that: \( 2 \tan^{-1} \left( \frac{1}{5} \right) + \sec^{-1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{-1} \left( \frac{1}{8} \right) = \frac{\pi}{4} \).
Answer: LHS \( = 2 \{ \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{8} \} + \sec^{-1} \frac{5\sqrt{2}}{7} \)
\( = 2 \tan^{-1} \left[ \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \cdot \frac{1}{8}} \right] + \tan^{-1} \sqrt{ \left( \frac{5\sqrt{2}}{7} \right)^2 - 1} \)
\( = 2 \tan^{-1} \frac{13/40}{39/40} + \tan^{-1} \sqrt{\frac{50}{49} - 1} \)
\( = 2 \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{7} \)
\( = \tan^{-1} \left[ \frac{2 \times \frac{1}{3}}{1 - (1/3)^2} \right] + \tan^{-1} \frac{1}{7} = \tan^{-1} \frac{2/3}{8/9} + \tan^{-1} \frac{1}{7} \)
\( = \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{1}{7} = \tan^{-1} \left[ \frac{3/4 + 1/7}{1 - \frac{3}{4} \cdot \frac{1}{7}} \right] = \tan^{-1} \frac{25/28}{25/28} = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS} \).

Question. If \( y = \cot^{-1}(\sqrt{\cos x}) - \tan^{-1}(\sqrt{\cos x}) \), then prove that \( \sin y = \tan^2 \left( \frac{x}{2} \right) \).
Answer: Given \( y = \cot^{-1}(\sqrt{\cos x}) - \tan^{-1}(\sqrt{\cos x}) \)
\( \because \tan^{-1} A + \cot^{-1} A = \frac{\pi}{2} \Rightarrow \cot^{-1} A = \frac{\pi}{2} - \tan^{-1} A \)
\( \therefore y = \frac{\pi}{2} - 2 \tan^{-1} (\sqrt{\cos x}) \)
\( \Rightarrow 2 \tan^{-1} (\sqrt{\cos x}) = \frac{\pi}{2} - y \)
\( \Rightarrow \cos^{-1} \left[ \frac{1 - (\sqrt{\cos x})^2}{1 + (\sqrt{\cos x})^2} \right] = \frac{\pi}{2} - y \)
\( \Rightarrow \frac{1 - \cos x}{1 + \cos x} = \cos \left( \frac{\pi}{2} - y \right) = \sin y \)
\( \Rightarrow \sin y = \frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} = \tan^2 \frac{x}{2} \).

Question. Prove that: \( \sin^{-1} \left( \frac{8}{17} \right) + \cos^{-1} \left( \frac{4}{5} \right) = \cot^{-1} \left( \frac{36}{77} \right) \).
Answer: Let \( \sin^{-1} \frac{8}{17} = \alpha \Rightarrow \sin \alpha = \frac{8}{17}, \cos \alpha = \frac{15}{17}, \cot \alpha = \frac{15}{8} \)
Let \( \cos^{-1} \frac{4}{5} = \beta \Rightarrow \cos \beta = \frac{4}{5}, \sin \beta = \frac{3}{5}, \cot \beta = \frac{4}{3} \)
Now \( \cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{\frac{15}{8} \cdot \frac{4}{3} - 1}{\frac{15}{8} + \frac{4}{3}} = \frac{\frac{60 - 24}{24}}{\frac{45 + 32}{24}} = \frac{36}{77} \)
\( \Rightarrow \alpha + \beta = \cot^{-1} \frac{36}{77} \)
\( \Rightarrow \sin^{-1} \frac{8}{17} + \cos^{-1} \frac{4}{5} = \cot^{-1} \frac{36}{77} \).

Question. Prove that: \( \cos^{-1} \left( \frac{4}{5} \right) + \cos^{-1} \left( \frac{12}{13} \right) = \cos^{-1} \left( \frac{33}{65} \right) \).
Answer: Let \( \cos^{-1} \frac{4}{5} = x, \cos^{-1} \frac{12}{13} = y \)
\( \Rightarrow \cos x = 4/5, \cos y = 12/13, \sin x = 3/5, \sin y = 5/13 \)
Now, \( \cos(x + y) = \cos x \cos y - \sin x \sin y \)
\( = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{48 - 15}{65} = \frac{33}{65} \)
\( \Rightarrow x + y = \cos^{-1} \frac{33}{65} \).
Putting the values of \( x \) and \( y \), we get
\( \cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65} \).

Question. Prove that: \( \tan \left( \frac{\pi}{4} + \frac{1}{2} \cos^{-1} \frac{a}{b} \right) + \tan \left( \frac{\pi}{4} - \frac{1}{2} \cos^{-1} \frac{a}{b} \right) = \frac{2b}{a} \).
Answer: Let \( \frac{1}{2} \cos^{-1} \frac{a}{b} = x \Rightarrow \cos 2x = \frac{a}{b} \).
LHS \( = \tan \left( \frac{\pi}{4} + x \right) + \tan \left( \frac{\pi}{4} - x \right) = \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} \)
\( = \frac{(1 + \tan x)^2 + (1 - \tan x)^2}{1 - \tan^2 x} = \frac{1 + \tan^2 x + 2 \tan x + 1 + \tan^2 x - 2 \tan x}{1 - \tan^2 x} \)
\( = \frac{2(1 + \tan^2 x)}{1 - \tan^2 x} = \frac{2}{\frac{1 - \tan^2 x}{1 + \tan^2 x}} = \frac{2}{\cos 2x} \)
\( = \frac{2}{a/b} = \frac{2b}{a} = \text{RHS} \).

Question. Find the value of \( \sin\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) \).
Answer: Let \( \cos^{-1} \frac{4}{5} = \alpha \Rightarrow \cos \alpha = 4/5, \sin \alpha = 3/5 \)
and \( \tan^{-1} \frac{2}{3} = \beta \Rightarrow \tan \beta = 2/3 \Rightarrow \sin \beta = \frac{2}{\sqrt{13}}, \cos \beta = \frac{3}{\sqrt{13}} \)
\( \therefore \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\( = \frac{3}{5} \cdot \frac{3}{\sqrt{13}} + \frac{4}{5} \cdot \frac{2}{\sqrt{13}} = \frac{9}{5\sqrt{13}} + \frac{8}{5\sqrt{13}} = \frac{17}{5\sqrt{13}} \).

Question. If \( \cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha \) prove that \( \frac{x^2}{a^2} - 2\frac{xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha \). [
Answer: Given, \( \cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha \)
\( \Rightarrow \cos^{-1} \left[ \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right] = \alpha \) [\( \because \cos^{-1} x + \cos^{-1} y = \cos^{-1} \{xy - \sqrt{1-x^2}\sqrt{1-y^2}\} \)]
\( \Rightarrow \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} = \cos \alpha \)
\( \Rightarrow \frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2}} \)
Squaring both sides,
\( \Rightarrow \left( \frac{xy}{ab} - \cos \alpha \right)^2 = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \)
\( \Rightarrow \frac{x^2 y^2}{a^2 b^2} + \cos^2 \alpha - 2 \frac{xy}{ab} \cos \alpha = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \)
\( \Rightarrow \frac{x^2}{a^2} - 2 \frac{xy}{ab} \cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha \)
\( \Rightarrow \frac{x^2}{a^2} - 2 \frac{xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha \). Hence proved.

Question. Prove that: \( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) = \frac{\pi}{4} - \frac{x}{2}, x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
Answer: \( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) = \tan^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2}} \right] \)
\( = \tan^{-1} \left[ \frac{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)}{\left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2} \right] \)
\( = \tan^{-1} \left[ \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \right] \)
\( = \tan^{-1} \left[ \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \right] \) [Divide each term by \( \cos \frac{x}{2} \)]
\( = \tan^{-1} \left[ \frac{\tan \frac{\pi}{4} - \tan \frac{x}{2}}{1 + \tan \frac{\pi}{4} \tan \frac{x}{2}} \right] \)
\( = \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - \frac{x}{2} \right) \right] = \frac{\pi}{4} - \frac{x}{2} \)
[\( \because x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \Rightarrow -\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{4} \Rightarrow \frac{\pi}{4} - \frac{x}{2} \in (0, \pi/2) \subseteq (-\pi/2, \pi/2) \)]

Question. Solve: \( \tan^{-1} (x - 1) + \tan^{-1} x + \tan^{-1} (x + 1) = \tan^{-1} 3x \) 
Answer: Given: \( \tan^{-1} (x - 1) + \tan^{-1} x + \tan^{-1} (x + 1) = \tan^{-1} 3x \)
\( \Rightarrow \tan^{-1} (x - 1) + \tan^{-1} (x + 1) = \tan^{-1} 3x - \tan^{-1} x \)
\( \Rightarrow \tan^{-1} \left[ \frac{(x - 1) + (x + 1)}{1 - (x - 1)(x + 1)} \right] = \tan^{-1} \left[ \frac{3x - x}{1 + 3x^2} \right] \)
\( \Rightarrow \tan^{-1} \left( \frac{2x}{1 - (x^2 - 1)} \right) = \tan^{-1} \left( \frac{2x}{1 + 3x^2} \right) \)
\( \Rightarrow \frac{2x}{2 - x^2} = \frac{2x}{1 + 3x^2} \)
Either \( x = 0 \) or \( 2 - x^2 = 1 + 3x^2 \Rightarrow 4x^2 = 1 \)
\( \Rightarrow x^2 = \frac{1}{4} \therefore x = \pm \frac{1}{2} \)
\( \therefore x = 0, \pm \frac{1}{2} \)

Question. If \( 0 < x < 1 \), then solve the following for \( x \): \( \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1} \left( \frac{8}{31} \right) \) 
Answer: Given \( \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1} \left( \frac{8}{31} \right) \) [\( \because 0 < x < 1 \Rightarrow (x + 1)(x - 1) < 1 \)]
\( \Rightarrow \tan^{-1} \left[ \frac{x + 1 + x - 1}{1 - (x + 1)(x - 1)} \right] = \tan^{-1} \frac{8}{31} \)
\( \Rightarrow \tan^{-1} \left( \frac{2x}{1 - (x^2 - 1)} \right) = \tan^{-1} \frac{8}{31} \Rightarrow \frac{2x}{2 - x^2} = \frac{8}{31} \)
\( \Rightarrow 62x = 16 - 8x^2 \Rightarrow 8x^2 + 62x - 16 = 0 \)
\( \Rightarrow 4x^2 + 31x - 8 = 0 \Rightarrow 4x^2 + 32x - x - 8 = 0 \)
\( \Rightarrow 4x(x + 8) - 1(x + 8) = 0 \Rightarrow (x + 8)(4x - 1) = 0 \)
\( \Rightarrow x = -8 \) or \( x = \frac{1}{4} \)
[\( x = -8 \) is not acceptable as \( 0 < x < 1 \)]
\( \therefore x = \frac{1}{4} \)

Question. Solve: \( \cos(\tan^{-1} x) = \sin(\cot^{-1} \frac{3}{4}) \) 
Answer: Given \( \cos(\tan^{-1} x) = \sin(\cot^{-1} \frac{3}{4}) \)
\( \Rightarrow \cos(\tan^{-1} x) = \cos \left( \frac{\pi}{2} - \cot^{-1} \frac{3}{4} \right) \)
\( \Rightarrow \tan^{-1} x = \frac{\pi}{2} - \cot^{-1} \frac{3}{4} \)
[\( \because \sin \theta = \cos(\pi/2 - \theta) \) and \( \tan^{-1} x + \cot^{-1} x = \pi/2 \)]
\( \Rightarrow \tan^{-1} x = \tan^{-1} \frac{3}{4} \)
\( \Rightarrow x = \frac{3}{4} \)

Question. Prove that: \( \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{5} \right) + \tan^{-1} \left( \frac{1}{8} \right) = \frac{\pi}{4} \) 
Answer: LHS \( = \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{5} \right) + \tan^{-1} \left( \frac{1}{8} \right) \)
\( = \tan^{-1} \left[ \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}} \right] + \tan^{-1} \frac{1}{8} \) [\( \because \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} < 1 \)]
\( = \tan^{-1} \left( \frac{7/10}{9/10} \right) + \tan^{-1} \frac{1}{8} = \tan^{-1} \frac{7}{9} + \tan^{-1} \frac{1}{8} \)
\( = \tan^{-1} \left[ \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \cdot \frac{1}{8}} \right] = \tan^{-1} \left[ \frac{\frac{56 + 9}{72}}{\frac{72 - 7}{72}} \right] \)
\( = \tan^{-1} \left( \frac{65}{65} \right) = \tan^{-1} (1) = \frac{\pi}{4} = \text{RHS} \)

Question. Prove that \( \sin^{-1} \left( \frac{4}{5} \right) + \tan^{-1} \left( \frac{5}{12} \right) + \cos^{-1} \left( \frac{63}{65} \right) = \frac{\pi}{2} \) 
Answer: We have to prove that \( \sin^{-1} \left( \frac{4}{5} \right) + \tan^{-1} \left( \frac{5}{12} \right) + \cos^{-1} \left( \frac{63}{65} \right) = \frac{\pi}{2} \)
i.e. \( \sin^{-1} \left( \frac{4}{5} \right) + \tan^{-1} \left( \frac{5}{12} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{63}{65} \right) = \sin^{-1} \left( \frac{63}{65} \right) \)
LHS \( = \sin^{-1} \left( \frac{4}{5} \right) + \tan^{-1} \left( \frac{5}{12} \right) \)
\( = \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{\frac{5}{12}}{\sqrt{1 + \left( \frac{5}{12} \right)^2}} \right) \) [\( \because \tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right) \)]
\( = \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5/12}{13/12} \right) = \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) \)
\( = \sin^{-1} \left[ \frac{4}{5} \sqrt{1 - \left( \frac{5}{13} \right)^2} + \frac{5}{13} \sqrt{1 - \left( \frac{4}{5} \right)^2} \right] \)
\( = \sin^{-1} \left[ \frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5} \right] = \sin^{-1} \left( \frac{48}{65} + \frac{15}{65} \right) = \sin^{-1} \left( \frac{63}{65} \right) = \text{RHS} \). Hence proved.

Question. Solve the following for \( x \): \( \tan^{-1} \left( \frac{x - 2}{x - 3} \right) + \tan^{-1} \left( \frac{x + 2}{x + 3} \right) = \frac{\pi}{4}, |x| < 1 \). 
Answer: Given: \( \tan^{-1} \left( \frac{x - 2}{x - 3} \right) + \tan^{-1} \left( \frac{x + 2}{x + 3} \right) = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1} \left[ \frac{\frac{x - 2}{x - 3} + \frac{x + 2}{x + 3}}{1 - \left( \frac{x - 2}{x - 3} \right) \left( \frac{x + 2}{x + 3} \right)} \right] = \frac{\pi}{4} \) [\( \because \frac{x - 2}{x - 3} \cdot \frac{x + 2}{x + 3} = \frac{x^2 - 4}{x^2 - 9} < 1 \text{ for } |x| < 1 \)]
\( \Rightarrow \tan^{-1} \left[ \frac{(x - 2)(x + 3) + (x + 2)(x - 3)}{(x - 3)(x + 3) - (x - 2)(x + 2)} \right] = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1} \left[ \frac{x^2 + 3x - 2x - 6 + x^2 - 3x + 2x - 6}{x^2 - 9 - x^2 + 4} \right] = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1} \left[ \frac{2x^2 - 12}{-5} \right] = \frac{\pi}{4} \Rightarrow \frac{2x^2 - 12}{-5} = \tan \frac{\pi}{4} = 1 \)
\( \Rightarrow 2x^2 - 12 = -5 \Rightarrow 2x^2 = 7 \Rightarrow x^2 = \frac{7}{2} \)
\( \Rightarrow x = \pm \sqrt{\frac{7}{2}} \), not acceptable as \( |x| < 1 \).
Hence, there is no solution.

Question. If \( (\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8} \), then find \( x \). 
Answer: Here, \( (\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8} \)
\( \Rightarrow (\tan^{-1} x)^2 + \left( \frac{\pi}{2} - \tan^{-1} x \right)^2 = \frac{5\pi^2}{8} \)
\( \Rightarrow (\tan^{-1} x)^2 + (\tan^{-1} x)^2 + \frac{\pi^2}{4} - \pi \tan^{-1} x = \frac{5\pi^2}{8} \)
\( \Rightarrow 2(\tan^{-1} x)^2 - \pi \tan^{-1} x + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0 \)
\( \Rightarrow 2(\tan^{-1} x)^2 - \pi \tan^{-1} x - \frac{3\pi^2}{8} = 0 \) ...(i)
Let \( \tan^{-1} x = y \), then (i) becomes
\( 2y^2 - \pi y - \frac{3\pi^2}{8} = 0 \Rightarrow 16y^2 - 8\pi y - 3\pi^2 = 0 \)
\( \Rightarrow 16y^2 - 12\pi y + 4\pi y - 3\pi^2 = 0 \Rightarrow 4y(4y - 3\pi) + \pi(4y - 3\pi) = 0 \)
\( \Rightarrow (4y - 3\pi)(4y + \pi) = 0 \Rightarrow y = \frac{3\pi}{4} \text{ or } y = -\frac{\pi}{4} \)
\( \tan^{-1} x = \frac{3\pi}{4} \) does not belong to domain of \( \tan^{-1} x \) i.e., \( (-\pi/2, \pi/2) \)
\( \therefore \tan^{-1} x = -\frac{\pi}{4} \Rightarrow x = \tan \left( -\frac{\pi}{4} \right) = -1 \)

Question. If \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \), \( x, y, z > 0 \), then find the value of \( xy + yz + zx \). 
Answer: Given \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \Rightarrow \tan^{-1} x + \tan^{-1} y = \frac{\pi}{2} - \tan^{-1} z \)
\( \Rightarrow \tan^{-1} \left( \frac{x + y}{1 - xy} \right) = \cot^{-1} z = \tan^{-1} \frac{1}{z} \)
\( \Rightarrow \frac{x + y}{1 - xy} = \frac{1}{z} \Rightarrow xz + yz = 1 - xy \Rightarrow xy + yz + zx = 1 \)

Question. Solve the equation for \( x \): \( \sin^{-1} x + \sin^{-1} (1 - x) = \cos^{-1} x \) 
Answer: \( \sin^{-1} x + \sin^{-1} (1 - x) = \cos^{-1} x \)
\( \Rightarrow \sin^{-1} \left[ x\sqrt{1 - (1 - x)^2} + (1 - x)\sqrt{1 - x^2} \right] = \sin^{-1} \sqrt{1 - x^2} \)
[\( \because \sin^{-1} x + \sin^{-1} y = \sin^{-1} \{x\sqrt{1-y^2} + y\sqrt{1-x^2}\} \) and \( \cos^{-1} x = \sin^{-1} \sqrt{1-x^2} \)]
\( \Rightarrow x\sqrt{1 - (1 + x^2 - 2x)} + (1 - x)\sqrt{1 - x^2} = \sqrt{1 - x^2} \)
\( \Rightarrow x\sqrt{2x - x^2} + \sqrt{1 - x^2} - x\sqrt{1 - x^2} = \sqrt{1 - x^2} \)
\( \Rightarrow x\sqrt{2x - x^2} - x\sqrt{1 - x^2} = 0 \Rightarrow x \{ \sqrt{2x - x^2} - \sqrt{1 - x^2} \} = 0 \)
\( \Rightarrow x = 0 \) or \( \sqrt{2x - x^2} - \sqrt{1 - x^2} = 0 \Rightarrow \sqrt{2x - x^2} = \sqrt{1 - x^2} \)
Now, \( 2x - x^2 = 1 - x^2 \)
Squaring both sides, we get
\( 2x - x^2 = 1 - x^2 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \)
\( \therefore 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \)
Hence, \( x = 0 \) and \( x = \frac{1}{2} \).

Question. Find the value of \( \cot \left\{ \frac{1}{2} \left[ \cos^{-1} \frac{2x}{1+x^2} + \sin^{-1} \frac{1-y^2}{1+y^2} \right] \right\} \), \( |x| < 1, y > 0 \text{ and } xy < 1 \). 
Answer: \( \cot \left\{ \frac{1}{2} \left[ \cos^{-1} \frac{2x}{1+x^2} + \sin^{-1} \frac{1-y^2}{1+y^2} \right] \right\} \)
\( = \cot \left\{ \frac{1}{2} \left[ \frac{\pi}{2} - \sin^{-1} \frac{2x}{1+x^2} + \frac{\pi}{2} - \cos^{-1} \frac{1-y^2}{1+y^2} \right] \right\} \)
\( = \cot \left\{ \frac{1}{2} \left[ \pi - 2\tan^{-1} x - 2\tan^{-1} y \right] \right\} \)
\( = \cot \left[ \frac{\pi}{2} - (\tan^{-1} x + \tan^{-1} y) \right] \)
\( = \tan(\tan^{-1} x + \tan^{-1} y) = \tan \left[ \tan^{-1} \left( \frac{x+y}{1-xy} \right) \right] \)
\( = \frac{x+y}{1-xy} \) [\( \because xy < 1 \)]

Question. Does the following trigonometric equation have any solutions? If yes, obtain the solution(s):
\( \tan^{-1} \left( \frac{x+1}{x-1} \right) + \tan^{-1} \left( \frac{x-1}{x} \right) = -\tan^{-1} 7 \). 
Answer: \( \tan^{-1} \left( \frac{x+1}{x-1} \right) + \tan^{-1} \left( \frac{x-1}{x} \right) = -\tan^{-1} 7 \)
\( \Rightarrow \tan^{-1} \left[ \frac{\left( \frac{x+1}{x-1} \right) + \left( \frac{x-1}{x} \right)}{1 - \left( \frac{x+1}{x-1} \right) \left( \frac{x-1}{x} \right)} \right] = -\tan^{-1} 7 \), if \( \left( \frac{x+1}{x-1} \right) \left( \frac{x-1}{x} \right) < 1 \)
\( \Rightarrow \tan^{-1} \left[ \frac{x(x+1) + (x-1)^2}{x(x-1) - (x+1)(x-1)} \right] = -\tan^{-1} 7 \)
\( \Rightarrow \frac{(x^2 + x) + (x^2 + 1 - 2x)}{(x^2 - x) - (x^2 - 1)} = \tan(\tan^{-1}(-7)) = -7 \)
\( \Rightarrow \frac{2x^2 - x + 1}{-x + 1} = -7 \Rightarrow 2x^2 - x + 1 = 7x - 7 \Rightarrow 2x^2 - 8x + 8 = 0 \)
\( \Rightarrow 2(x^2 - 4x + 4) = 0 \Rightarrow (x-2)^2 = 0 \Rightarrow x = 2 \)
Let us now verify whether \( x = 2 \) satisfies the condition \( \left( \frac{x+1}{x-1} \right) \left( \frac{x-1}{x} \right) < 1 \).
For \( x = 2 \), \( \left( \frac{2+1}{2-1} \right) \left( \frac{2-1}{2} \right) = 3 \times \frac{1}{2} = \frac{3}{2} \), which is not less than 1.
Hence, this value does not satisfy the condition.
i.e., there is no solution of the given trigonometric equation.

Question. Find the real solution of \( \tan^{-1} \sqrt{x(x+1)} + \sin^{-1} \sqrt{x^2+x+1} = \frac{\pi}{2} \). 
Answer: We have, \( \tan^{-1} \sqrt{x(x+1)} + \sin^{-1} \sqrt{x^2+x+1} = \frac{\pi}{2} \) ...(i)
Let \( \sin^{-1} \sqrt{x^2+x+1} = \theta \)
\( \Rightarrow \sin \theta = \sqrt{x^2+x+1} \Rightarrow \tan \theta = \frac{\sqrt{x^2+x+1}}{\sqrt{1 - (x^2+x+1)}} = \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} \)
\( \therefore \theta = \tan^{-1} \frac{\sqrt{x^2+x+1}}{\sqrt{-x(x+1)}} \)
On putting the value of \( \theta \) in equation (i), we get
\( \tan^{-1} \sqrt{x(x+1)} + \tan^{-1} \frac{\sqrt{x^2+x+1}}{\sqrt{-x(x+1)}} = \frac{\pi}{2} \)
\( \tan^{-1} \left[ \frac{\sqrt{x(x+1)} + \frac{\sqrt{x^2+x+1}}{\sqrt{-x(x+1)}}}{1 - \sqrt{x(x+1)} \cdot \frac{\sqrt{x^2+x+1}}{\sqrt{-x(x+1)}}} \right] = \frac{\pi}{2} \)
\( \Rightarrow \frac{x(x+1) + \sqrt{-(x^2+x+1)x(x+1)}}{ \sqrt{-x(x+1)} - \sqrt{x(x+1)} \sqrt{x^2+x+1} } = \tan \frac{\pi}{2} = \frac{1}{0} \)
\( \Rightarrow 1 - \sqrt{x(x+1)} \cdot \frac{\sqrt{x^2+x+1}}{i\sqrt{x(x+1)}} = 0 \) is complex.
Alternatively, for the expressions to be real:
\( x(x+1) \ge 0 \) and \( x^2+x+1 \le 1 \)
\( \Rightarrow x^2+x \ge 0 \) and \( x^2+x \le 0 \)
The only possibility is \( x^2+x = 0 \)
\( \Rightarrow x(x+1) = 0 \Rightarrow x = 0 \) or \( x = -1 \).
Both values satisfy the original equation.
For real solution, we have \( x = 0, -1 \).

Question. If \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \), then prove that: \( x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz \). 
Answer: Let \( \sin^{-1} x = A \Rightarrow \sin A = x \)
\( \sin^{-1} y = B \Rightarrow \sin B = y \)
\( \sin^{-1} z = C \Rightarrow \sin C = z \)
Given, \( A + B + C = \pi \Rightarrow 2A + 2B + 2C = 2\pi \)
\( \therefore \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \) [Using trigonometric property]
\( \Rightarrow 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C = 4 \sin A \sin B \sin C \)
\( \Rightarrow 2 \sin A \sqrt{1 - \sin^2 A} + 2 \sin B \sqrt{1 - \sin^2 B} + 2 \sin C \sqrt{1 - \sin^2 C} = 4 \sin A \sin B \sin C \)
\( \Rightarrow 2x\sqrt{1-x^2} + 2y\sqrt{1-y^2} + 2z\sqrt{1-z^2} = 4xyz \)
\( \Rightarrow x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz \). Hence proved.

Question. If \( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \), then prove that \( a+b+c = abc \). 
Answer: Let \( \tan^{-1} a = \alpha \Rightarrow \tan \alpha = a \)
\( \tan^{-1} b = \beta \Rightarrow \tan \beta = b \)
\( \tan^{-1} c = \gamma \Rightarrow \tan \gamma = c \)
Now, given \( \alpha + \beta + \gamma = \pi \Rightarrow \alpha + \beta = \pi - \gamma \)
Taking tangent on both sides,
\( \tan(\alpha + \beta) = \tan(\pi - \gamma) \)
\( \Rightarrow \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = -\tan \gamma \)
\( \Rightarrow \tan \alpha + \tan \beta = -\tan \gamma (1 - \tan \alpha \tan \beta) \)
\( \Rightarrow \tan \alpha + \tan \beta = -\tan \gamma + \tan \alpha \tan \beta \tan \gamma \)
\( \Rightarrow \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \)
Thus, \( a + b + c = abc \). Hence proved.

Question. Show that: \( 2\tan^{-1} \left\{ \tan \frac{\alpha}{2} \tan \left( \frac{\pi}{4} - \frac{\beta}{2} \right) \right\} = \tan^{-1} \frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta} \).
Answer: LHS \( = 2\tan^{-1} \left\{ \tan \frac{\alpha}{2} \tan \left( \frac{\pi}{4} - \frac{\beta}{2} \right) \right\} \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} \tan \left( \frac{\pi}{4} - \frac{\beta}{2} \right)}{1 - \tan^2 \frac{\alpha}{2} \tan^2 \left( \frac{\pi}{4} - \frac{\beta}{2} \right)} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} \cdot \left( \frac{1 - \tan \frac{\beta}{2}}{1 + \tan \frac{\beta}{2}} \right)}{1 - \tan^2 \frac{\alpha}{2} \cdot \left( \frac{1 - \tan \frac{\beta}{2}}{1 + \tan \frac{\beta}{2}} \right)^2} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} (1 - \tan \frac{\beta}{2})(1 + \tan \frac{\beta}{2})}{(1 + \tan \frac{\beta}{2})^2 - \tan^2 \frac{\alpha}{2} (1 - \tan \frac{\beta}{2})^2} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} (1 - \tan^2 \frac{\beta}{2})}{(1 + \tan^2 \frac{\beta}{2} + 2 \tan \frac{\beta}{2}) - \tan^2 \frac{\alpha}{2} (1 + \tan^2 \frac{\beta}{2} - 2 \tan \frac{\beta}{2})} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} (1 - \tan^2 \frac{\beta}{2})}{(1 + \tan^2 \frac{\beta}{2})(1 - \tan^2 \frac{\alpha}{2}) + 2 \tan \frac{\beta}{2}(1 + \tan^2 \frac{\alpha}{2})} \right] \)
Dividing \( N^r \) and \( D^r \) by \( (1 + \tan^2 \frac{\alpha}{2})(1 + \tan^2 \frac{\beta}{2}) \):
\( = \tan^{-1} \left[ \frac{ \frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \cdot \frac{1 - \tan^2 \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}} }{ \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} + \frac{2 \tan \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}} } \right] \)
\( = \tan^{-1} \left( \frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta} \right) = \text{RHS} \)

Question. If \( a_1, a_2, a_3, \dots, a_n \) is an arithmetic progression with common difference \( d \), then evaluate the following expression. 
\( \tan \left[ \tan^{-1} \left( \frac{d}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{d}{1 + a_2 a_3} \right) + \tan^{-1} \left( \frac{d}{1 + a_3 a_4} \right) + \dots + \tan^{-1} \left( \frac{d}{1 + a_{n-1} a_n} \right) \right] \).
Answer: We have, \( d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1} \)
Given expression is \( \tan \left[ \tan^{-1} \left( \frac{a_2 - a_1}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{a_3 - a_2}{1 + a_2 a_3} \right) + \dots + \tan^{-1} \left( \frac{a_n - a_{n-1}}{1 + a_{n-1} a_n} \right) \right] \)
\( = \tan [ (\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + \dots + (\tan^{-1} a_n - \tan^{-1} a_{n-1}) ] \)
\( = \tan [\tan^{-1} a_n - \tan^{-1} a_1] \)
\( = \tan \left[ \tan^{-1} \left( \frac{a_n - a_1}{1 + a_n a_1} \right) \right] \)
\( = \frac{a_n - a_1}{1 + a_n a_1} \)

PROFICIENCY EXERCISE

Objective Type Questions:

Question. Which of the following corresponds to the principal value branch of \( \tan^{-1} x \)?
(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) \( (0, \pi) \)
Answer: (a)

Question. The value of \( \tan^{-1} (1) + \cos^{-1} \left( -\frac{1}{2} \right) + \sin^{-1} \left( -\frac{1}{2} \right) \) corresponding to principal branches is:
(a) \( \frac{3\pi}{4} \)
(b) \( \frac{\pi}{4} \)
(c) \( -\frac{\pi}{4} \)
(d) \( -\frac{3\pi}{4} \)
Answer: (a)

Question. The principal value of \( \sin^{-1} \left( \frac{-\sqrt{3}}{2} \right) \) is:
(a) \( -\frac{2\pi}{3} \)
(b) \( -\frac{\pi}{3} \)
(c) \( \frac{4\pi}{3} \)
(d) \( \frac{5\pi}{3} \)
Answer: (b)

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

Question. If \( \tan^{-1} x = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} \), then \( x \) is equal to
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \pi \)
Answer: (a)

Question. If \( \tan^{-1} x + 2 \cot^{-1} x = \pi \), then \( x \) equals
(a) 0
(b) 1
(c) -1
(d) \( \frac{1}{2} \)
Answer: (a)

Question. Which of the following is the principal value branch of \( \csc^{-1} x \)?
(a) \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
(b) \( [0, \pi] - \left\{ \frac{\pi}{2} \right\} \)
(c) \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)
(d) \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \)
Answer: (d)

Question. If \( \cos^{-1} x + \cos^{-1} y = \frac{\pi}{2} \), then the value of \( \sin^{-1} x + \sin^{-1} y \) is
(a) 0
(b) \( \frac{\pi}{2} \)
(c) \( \pi \)
(d) \( \frac{2\pi}{3} \)
Answer: (b)