CBSE Class 12 Mathematics Inverse Trigonometric Functions Worksheet Set 03

Very Short Answer Questions [1 mark]

Question. Write the principal value of \( \tan^{-1} 1 + \cos^{-1}\left( -\frac{1}{2} \right) \).
Answer: \( \tan^{-1} 1 + \cos^{-1}\left( -\frac{1}{2} \right) = \tan^{-1}\left( \tan \frac{\pi}{4} \right) + \cos^{-1}\left( \cos \left( \pi - \frac{\pi}{3} \right) \right) \)
\( = \tan^{-1}\left( \tan \frac{\pi}{4} \right) + \cos^{-1}\left( \cos \frac{2\pi}{3} \right) \)
\( = \frac{\pi}{4} + \frac{2\pi}{3} \) \( \left[ \because \frac{\pi}{4} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \text{ and } \frac{2\pi}{3} \in [0, \pi] \right] \)
\( = \frac{3\pi + 8\pi}{12} = \frac{11\pi}{12} \)

Question. Find the value of \( \sin\left[ \frac{\pi}{3} - \sin^{-1}\left( -\frac{1}{2} \right) \right] \).
Answer: \( \sin\left[ \frac{\pi}{3} - \sin^{-1}\left( -\frac{1}{2} \right) \right] = \sin\left[ \frac{\pi}{3} + \sin^{-1}\left( \frac{1}{2} \right) \right] \) \( [\because \sin^{-1}(-x) = -\sin^{-1} x] \)
\( = \sin\left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \sin\left( \frac{\pi}{2} \right) = 1 \)

Question. Write the value of \( \tan\left( 2 \tan^{-1} \frac{1}{5} \right) \).
Answer: \( \tan\left( 2 \tan^{-1} \frac{1}{5} \right) = \tan\left[ \tan^{-1}\left( \frac{2 \times \frac{1}{5}}{1 - \left( \frac{1}{5} \right)^2} \right) \right] \) \( \left[ \because 2 \tan^{-1} x = \tan^{-1}\left( \frac{2x}{1 - x^2} \right) \right] \)
\( = \tan\left[ \tan^{-1}\left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) \right] = \tan\left[ \tan^{-1}\left( \frac{2}{5} \times \frac{25}{24} \right) \right] \)
\( = \tan\left( \tan^{-1} \frac{5}{12} \right) = \frac{5}{12} \)

Question. Write the principal value of \( \cos^{-1}\left( \cos \frac{7\pi}{6} \right) \).
Answer: \( \cos^{-1}\left( \cos \frac{7\pi}{6} \right) = \cos^{-1}\left( \cos\left( 2\pi - \frac{5\pi}{6} \right) \right) \)
\( = \cos^{-1}\left( \cos \frac{5\pi}{6} \right) = \frac{5\pi}{6} \) \( \left[ \because \frac{5\pi}{6} \in [0, \pi] \right] \)

Question. What is the principal value of \( \sin^{-1}\left( -\frac{\sqrt{3}}{2} \right) \)?
Answer: \( \sin^{-1}\left( -\frac{\sqrt{3}}{2} \right) = \sin^{-1}\left( -\sin \frac{\pi}{3} \right) \) \( \left[ \because \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \right] \)
\( = \sin^{-1}\left( \sin\left( -\frac{\pi}{3} \right) \right) = -\frac{\pi}{3} \) \( \left[ \because -\frac{\pi}{3} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \right] \)

Question. Find the value of \( \sin^{-1}\left( \sin \frac{4\pi}{5} \right) \).
Answer: We are given \( \sin^{-1}\left( \sin \frac{4\pi}{5} \right) = \sin^{-1}\left( \sin\left( \pi - \frac{\pi}{5} \right) \right) = \sin^{-1}\left( \sin \frac{\pi}{5} \right) = \frac{\pi}{5} \)

Question. Write the principal value of \( \tan^{-1}(-1) \).
Answer: \( \tan^{-1}(-1) = \tan^{-1}\left( -\tan \frac{\pi}{4} \right) \)
\( = \tan^{-1}\left( \tan\left( -\frac{\pi}{4} \right) \right) = -\frac{\pi}{4} \) \( \left[ \because -\frac{\pi}{4} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \right] \) range of the principal value branch of \( \tan^{-1} \) function and \( \tan\left( -\frac{\pi}{4} \right) = -1 \)

Question. Write the principal value of \( \cos^{-1}\left( \frac{1}{2} \right) + 2 \sin^{-1}\left( \frac{1}{2} \right) \).
Answer: We have, \( \cos^{-1}\left( \frac{1}{2} \right) = \cos^{-1}\left( \cos \frac{\pi}{3} \right) \)
\( = \frac{\pi}{3} \) \( \left[ \because \frac{\pi}{3} \in [0, \pi] \right] \)
Also, \( \sin^{-1}\left( \frac{1}{2} \right) = \sin^{-1}\left( \sin \frac{\pi}{6} \right) \)
\( = \frac{\pi}{6} \) \( \left[ \because \frac{\pi}{6} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \right] \)
\( \therefore \cos^{-1}\left( \frac{1}{2} \right) + 2 \sin^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{3} + 2\left( \frac{\pi}{6} \right) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \)
[Note: Principal value branches of \( \sin x \) and \( \cos x \) are \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \) and \( [0, \pi] \) respectively.]

Question. Write the value of \( \cot(\tan^{-1} a + \cot^{-1} a) \).
Answer: \( \cot(\tan^{-1} a + \cot^{-1} a) = \cot \frac{\pi}{2} = 0 \) \( [\because \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \forall x \in \mathbb{R}] \)

Question. Write the value of \( \sin\left( 2 \sin^{-1} \frac{3}{5} \right) \).
Answer: Let \( \sin\left( 2 \sin^{-1} \frac{3}{5} \right) = \theta \)
\( \implies \) \( 2 \sin^{-1} \frac{3}{5} = \sin^{-1} \theta \)
\( \implies \) \( \sin^{-1}\left\{ 2 \times \frac{3}{5} \sqrt{1 - \frac{9}{25}} \right\} = \sin^{-1} \theta \) \( [\because 2 \sin^{-1} x = \sin^{-1}(2x\sqrt{1-x^2})] \)
\( \implies \) \( \sin^{-1}\left\{ \frac{6}{5} \times \frac{4}{5} \right\} = \sin^{-1} \theta \)
\( \implies \) \( \sin^{-1}\left( \frac{24}{25} \right) = \sin^{-1} \theta \)
\( \implies \) \( \theta = \frac{24}{25} \)
\( \implies \) \( \sin\left( 2 \sin^{-1} \frac{3}{5} \right) = \frac{24}{25} \)

Question. Write the principal value of \( \tan^{-1}\left( \tan \frac{7\pi}{6} \right) \).
Answer: \( \tan^{-1}\left( \tan \frac{7\pi}{6} \right) = \tan^{-1}\left( \tan\left( \pi + \frac{\pi}{6} \right) \right) \)
\( = \tan^{-1}\left( \tan \frac{\pi}{6} \right) = \frac{\pi}{6} \) \( \left[ \because \frac{\pi}{6} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \right] \)

Question. If \( \sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \), then find the value of \( x \).
Answer: Given \( \sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \)
\( \implies \) \( \sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1} 1 \)
\( \implies \) \( \sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2} \)
\( \implies \) \( \sin^{-1} \frac{1}{5} = \frac{\pi}{2} - \cos^{-1} x \)
\( \implies \) \( \sin^{-1} \frac{1}{5} = \sin^{-1} x \)
\( \implies \) \( x = \frac{1}{5} \)

Question. Evaluate: \( \tan(\tan^{-1}(-4)) \).
Answer: \( \tan(\tan^{-1}(-4)) = -4 \) \( [\because \tan(\tan^{-1} x) = x \text{ if } x \in \mathbb{R} \text{ and } -4 \in \mathbb{R}] \)

Question. Find the value of \( \sin^{-1}\left( \cos \frac{43\pi}{5} \right) \).
Answer: \( \sin^{-1}\left( \cos \frac{43\pi}{5} \right) = \sin^{-1}\left( \cos\left( 8\pi + \frac{3\pi}{5} \right) \right) = \sin^{-1}\left( \cos \frac{3\pi}{5} \right) = \sin^{-1}\left( \sin\left( \frac{\pi}{2} - \frac{3\pi}{5} \right) \right) \)
\( = \sin^{-1}\left( \sin\left( -\frac{\pi}{10} \right) \right) = -\frac{\pi}{10} \) \( \left[ \because -\frac{\pi}{10} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \right] \)

Question. Find the principal value of \( \cos^{-1}[\cos(-680^\circ)] \).
Answer: \( \cos^{-1}[\cos(-680^\circ)] = \cos^{-1}[\cos 680^\circ] \) \( [\because \cos(-\theta) = \cos \theta] \)
\( = \cos^{-1}[\cos(720^\circ - 40^\circ)] = \cos^{-1}[\cos(4\pi - 40^\circ)] = \cos^{-1}(\cos 40^\circ) \)
\( = 40^\circ \text{ or } \frac{2\pi}{9} \) \( \left[ \because 40^\circ = \frac{2\pi}{9} \in [0, \pi] \right] \)

Short Answer Questions [2 marks]

Question. Write \( \cot^{-1}\left( \frac{1}{\sqrt{x^2 - 1}} \right) \), \( |x| > 1 \) in simplest form.
Answer: \( \cot^{-1}\left( \frac{1}{\sqrt{x^2 - 1}} \right) \)
Let \( x = \sec \theta \)
\( \implies \) \( \theta = \sec^{-1} x \)
Now, \( \cot^{-1}\left( \frac{1}{\sqrt{x^2 - 1}} \right) = \cot^{-1}\left( \frac{1}{\sqrt{\sec^2 \theta - 1}} \right) \)
\( = \cot^{-1}\left( \frac{1}{\tan \theta} \right) = \cot^{-1}(\cot \theta) = \theta = \sec^{-1} x \)

Question. Write the principal value of \( \tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) \).
Answer: \( \tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3}) = \tan^{-1}\left( \tan \frac{\pi}{3} \right) - \cot^{-1}\left( -\cot \frac{\pi}{6} \right) \)
\( = \tan^{-1}\left( \tan \frac{\pi}{3} \right) - \cot^{-1}\left( \cot\left( \pi - \frac{\pi}{6} \right) \right) = \tan^{-1}\left( \tan \frac{\pi}{3} \right) - \cot^{-1}\left( \cot \frac{5\pi}{6} \right) \)
\( = \frac{\pi}{3} - \frac{5\pi}{6} \) \( \left[ \because \frac{\pi}{3} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \text{ and } \frac{5\pi}{6} \in (0, \pi) \right] \)
\( = \frac{2\pi - 5\pi}{6} = -\frac{\pi}{2} \)

Question. What is the principal value of \( \cos^{-1}\left( \cos \frac{2\pi}{3} \right) + \sin^{-1}\left( \sin \frac{2\pi}{3} \right) \)?
Answer: \( \cos^{-1}\left( \cos \frac{2\pi}{3} \right) + \sin^{-1}\left( \sin \frac{2\pi}{3} \right) = \cos^{-1}\left( \cos \frac{2\pi}{3} \right) + \sin^{-1}\left( \sin\left( \pi - \frac{\pi}{3} \right) \right) \) \( \left[ \because \frac{2\pi}{3} \notin \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \right] \)
\( = \cos^{-1}\left( \cos \frac{2\pi}{3} \right) + \sin^{-1}\left( \sin \frac{\pi}{3} \right) = \frac{2\pi}{3} + \frac{\pi}{3} \)
\( = \frac{3\pi}{3} = \pi \) \( [\because \sin^{-1}(\sin x) = x \text{ if } x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \text{ and } \cos^{-1}(\cos x) = x \text{ if } x \in [0, \pi]] \)

Question. Write the value of \( \tan^{-1}\left[ 2 \sin\left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right] \).
Answer: \( \tan^{-1}\left[ 2 \sin\left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right] = \tan^{-1}\left[ 2 \sin\left( 2 \times \frac{\pi}{6} \right) \right] \) \( \left[ \because \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6} \right] \)
\( = \tan^{-1}\left( 2 \sin \frac{\pi}{3} \right) = \tan^{-1}\left( 2 \times \frac{\sqrt{3}}{2} \right) \)
\( = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)

Question. Prove that: \( \tan^{-1}\sqrt{x} = \frac{1}{2} \cos^{-1}\left( \frac{1-x}{1+x} \right), x \in (0, 1) \).
Answer: RHS = \( \frac{1}{2} \cos^{-1}\left( \frac{1-x}{1+x} \right) \)
\( = \frac{1}{2} \cos^{-1}\left( \frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2} \right) \) \( [\because 0 < x < 1 \implies 0 < \sqrt{x} < 1 \implies \sqrt{x} \ge 0] \)
\( = \frac{1}{2} . 2 \tan^{-1}\sqrt{x} = \tan^{-1}\sqrt{x} = \text{LHS} \)

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

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

Question. If \( 2 \tan^{-1}(\cos \theta) = \tan^{-1}(2 \text{cosec } \theta) \), then show that \( \theta = \frac{\pi}{4} \).
Answer: We have, \( 2 \tan^{-1}(\cos \theta) = \tan^{-1}(2 \text{cosec } \theta) \)
\( \implies \) \( \tan^{-1}\left( \frac{2 \cos \theta}{1 - \cos^2 \theta} \right) = \tan^{-1}(2 \text{cosec } \theta) \) \( \left[ \because 2 \tan^{-1} x = \tan^{-1}\left( \frac{2x}{1-x^2} \right) \right] \)
\( \implies \) \( \left( \frac{2 \cos \theta}{\sin^2 \theta} \right) = 2 \text{cosec } \theta \)
\( \implies \) \( \cot \theta . 2 \text{cosec } \theta = 2 \text{cosec } \theta \)
\( \implies \) \( \cot \theta = 1 \)
\( \implies \) \( \cot \theta = \cot \frac{\pi}{4} \)
\( \implies \) \( \theta = \frac{\pi}{4} \)

Question. Find the value of \( 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} \).
Answer: We have, \( 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} = 2 . 2 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} \)
\( = 2 . \tan^{-1}\left[ \frac{\frac{2}{5}}{1 - \left( \frac{1}{5} \right)^2} \right] - \tan^{-1} \frac{1}{239} \) \( \left[ \because 2 \tan^{-1} x = \tan^{-1}\left( \frac{2x}{1-x^2} \right) \right] \)
\( = 2 . \tan^{-1}\left[ \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right] - \tan^{-1} \frac{1}{239} \)
\( = 2 . \tan^{-1}\left[ \frac{2/5}{24/25} \right] - \tan^{-1} \frac{1}{239} \)
\( = 2 \tan^{-1} \frac{5}{12} - \tan^{-1} \frac{1}{239} \)

Long Answer Questions-I [4 marks]

Question. Show that: \( \tan\left( \frac{1}{2} \sin^{-1} \frac{3}{4} \right) = \frac{4-\sqrt{7}}{3} \).
Answer: Let \( \sin^{-1} \frac{3}{4} = \theta \)
\( \implies \) \( \sin \theta = \frac{3}{4} \) \( \left[ \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \right] \)
\( \implies \) \( \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} = \frac{3}{4} \) \( \left[ \because \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \right] \)
\( \implies \) \( 3 + 3 \tan^2 \frac{\theta}{2} = 8 \tan \frac{\theta}{2} \)
\( \implies \) \( 3 \tan^2 \frac{\theta}{2} - 8 \tan \frac{\theta}{2} + 3 = 0 \)
\( \implies \) \( \tan \frac{\theta}{2} = \frac{8 \pm \sqrt{64-36}}{6} \)
\( \implies \) \( \tan \frac{\theta}{2} = \frac{8 \pm \sqrt{28}}{6} \)
\( \implies \) \( \tan \frac{\theta}{2} = \frac{8 \pm 2\sqrt{7}}{6} \)
\( \implies \) \( \tan \frac{\theta}{2} = \frac{4 \pm \sqrt{7}}{3} \)
\( \implies \) \( \tan\left( \frac{1}{2} \sin^{-1} \frac{3}{4} \right) = \frac{4-\sqrt{7}}{3} \) \( \left[ \because \theta = \sin^{-1} \frac{3}{4} \right] \)

Question. Find the value of \( \tan^{-1}\left( \frac{x}{y} \right) - \tan^{-1}\left( \frac{x-y}{x+y} \right) \).
Answer: \( \tan^{-1}\left( \frac{x}{y} \right) - \tan^{-1}\left( \frac{x-y}{x+y} \right) = \tan^{-1}\left[ \frac{\frac{x}{y} - \frac{x-y}{x+y}}{1 + \frac{x}{y} . \frac{x-y}{x+y}} \right] \) \( \left[ \text{Here } \frac{x}{y} . \frac{x-y}{x+y} > -1 \right] \)
\( = \tan^{-1}\left( \frac{\frac{x^2 + xy - xy + y^2}{y(x+y)}}{\frac{xy + y^2 + x^2 - xy}{y(x+y)}} \right) \)
\( = \tan^{-1}\left( \frac{x^2 + y^2}{x^2 + y^2} \right) = \tan^{-1}(1) = \frac{\pi}{4} \)

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

Question. Prove that: \( \cot^{-1} 7 + \cot^{-1} 8 + \cot^{-1} 18 = \cot^{-1} 3 \).
Answer: LHS = \( \cot^{-1} 7 + \cot^{-1} 8 + \cot^{-1} 18 \)
\( = \left( \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} \right) + \tan^{-1} \frac{1}{18} \)
\( = \tan^{-1}\left[ \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right] + \tan^{-1} \frac{1}{18} \) \( \left[ \because \frac{1}{7} \times \frac{1}{8} < 1 \right] \)
\( = \tan^{-1} \frac{3}{11} + \tan^{-1} \frac{1}{18} = \tan^{-1}\left[ \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \times \frac{1}{18}} \right] \) \( \left[ \because \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left( \frac{x+y}{1-xy} \right) \right] \)
\( = \tan^{-1}\left[ \frac{\frac{65}{198}}{\frac{195}{198}} \right] = \tan^{-1}\left( \frac{65}{195} \right) = \tan^{-1} \frac{1}{3} \) \( \left[ \because \frac{3}{11} \times \frac{1}{18} < 1 \right] \)
\( = \cot^{-1} 3 = \text{RHS} \)

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 \( \sin^{-1}\left( \frac{5}{13} \right) = \alpha, \cos^{-1}\left( \frac{3}{5} \right) = \beta \)
\( \implies \) \( \sin \alpha = \frac{5}{13}, \cos \beta = \frac{3}{5} \)
\( \implies \) \( \cos \alpha = \sqrt{1 - \left( \frac{5}{13} \right)^2}, \sin \beta = \sqrt{1 - \left( \frac{3}{5} \right)^2} \)
\( \implies \) \( \cos \alpha = \frac{12}{13}, \sin \beta = \frac{4}{5} \)
Now, \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\( = \frac{5}{13} . \frac{3}{5} + \frac{12}{13} . \frac{4}{5} = \frac{15}{65} + \frac{48}{65} = \frac{63}{65} \)
\( \implies \) \( \alpha + \beta = \sin^{-1}\left( \frac{63}{65} \right) \)
Putting the value of \( \alpha \) and \( \beta \), we get
\( \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5} = \sin^{-1}\left( \frac{63}{65} \right) \)

Question. Prove the following: \( \cos[\tan^{-1} \{\sin(\cot^{-1} x)\}] = \sqrt{\frac{1+x^2}{2+x^2}} \).
Answer: LHS = \( \cos[\tan^{-1} \{\sin(\cot^{-1} x)\}] \)
Let \( \cot^{-1} x = \theta \)
\( \implies \) \( x = \cot \theta \)
\( = \cos[\tan^{-1}(\sin \theta)] = \cos\left[ \tan^{-1}\left( \frac{1}{\text{cosec } \theta} \right) \right] \)
\( = \cos\left[ \tan^{-1}\left( \frac{1}{\sqrt{1 + \cot^2 \theta}} \right) \right] = \cos\left[ \tan^{-1}\left( \frac{1}{\sqrt{1 + x^2}} \right) \right] \)
Let \( \tan^{-1} \frac{1}{\sqrt{1 + x^2}} = \alpha \)
\( \implies \) \( \frac{1}{\sqrt{1 + x^2}} = \tan \alpha \)
\( \implies \) \( \frac{1}{1 + x^2} = \tan^2 \alpha \)
\( \implies \) \( \frac{1}{1 + x^2} = \frac{\sin^2 \alpha}{\cos^2 \alpha} \)
\( \implies \) \( \frac{1}{1 + x^2} + 1 = \frac{\sin^2 \alpha}{\cos^2 \alpha} + 1 \)
\( \implies \) \( \frac{2 + x^2}{1 + x^2} = \frac{1}{\cos^2 \alpha} \)
\( \implies \) \( \cos \alpha = \frac{\sqrt{1 + x^2}}{\sqrt{2 + x^2}} \)
\( \implies \) \( \alpha = \cos^{-1}\left( \sqrt{\frac{1 + x^2}{2 + x^2}} \right) \)
\( \implies \) \( \cos \alpha = \cos\left( \cos^{-1} \sqrt{\frac{1 + x^2}{2 + x^2}} \right) = \sqrt{\frac{1 + x^2}{2 + x^2}} = \text{RHS} \)

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

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 \( y = \cot^{-1}(\sqrt{\cos x}) - \tan^{-1}(\sqrt{\cos x}) \)
\( \implies \) \( y = \frac{\pi}{2} - \tan^{-1}(\sqrt{\cos x}) - \tan^{-1}(\sqrt{\cos x}) \)
\( \implies \) \( y = \frac{\pi}{2} - 2 \tan^{-1}(\sqrt{\cos x}) \)
\( \implies \) \( y = \frac{\pi}{2} - \cos^{-1}\left( \frac{1 - \cos x}{1 + \cos x} \right) \)
\( \implies \) \( y = \sin^{-1}\left( \frac{1 - \cos x}{1 + \cos x} \right) \)
\( \implies \) \( \sin y = \frac{1 - \cos x}{1 + \cos x} \)
\( \implies \) \( \sin y = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \)
\( \implies \) \( \sin y = \tan^2 \frac{x}{2} \)
[Note: \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}, x \in \mathbb{R} \); \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, x \in [-1, 1] \) and \( 2 \tan^{-1} x = \cos^{-1} \frac{1 - x^2}{1 + x^2}, x \ge 0 \)]

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 \( \sin^{-1}\left( \frac{8}{17} \right) = \alpha \)
\( \implies \) \( \sin \alpha = \frac{8}{17} \)
\( \cos^{-1}\left( \frac{4}{5} \right) = \beta \)
\( \implies \) \( \cos \beta = \frac{4}{5} \)
\( \cos \alpha = \frac{15}{17} \implies \cot \alpha = \frac{15}{8} \), \( \sin \beta = \frac{3}{5} \implies \cot \beta = \frac{4}{3} \)
Now \( \cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{\frac{15}{8} . \frac{4}{3} - 1}{\frac{15}{8} + \frac{4}{3}} = \frac{\frac{60}{24} - 1}{\frac{45 + 32}{24}} = \frac{60 - 24}{45 + 32} = \frac{36}{77} \)
\( \therefore \alpha + \beta = \cot^{-1}\left( \frac{36}{77} \right) \)
\( \implies \sin^{-1}\left( \frac{8}{17} \right) + \cos^{-1}\left( \frac{4}{5} \right) = \cot^{-1}\left( \frac{36}{77} \right) \) Hence proved.

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: LHS \( = \text{Let } \cos^{-1} \frac{4}{5} = x, \cos^{-1} \frac{12}{13} = y \) \( [x, y \in [0, \pi]] \)
\( \implies \cos x = \frac{4}{5}, \cos y = \frac{12}{13} \)
\( \therefore \sin x = \sqrt{1 - \left( \frac{4}{5} \right)^2}, \sin y = \sqrt{1 - \left( \frac{12}{13} \right)^2} \) \( [\because x, y \in [0, \pi] \implies \sin x \text{ and } \sin y \text{ are +ve}] \)
\( \implies \sin x = \frac{3}{5}, \sin y = \frac{5}{13} \)
Now, \( \cos(x + y) = \cos x . \cos y - \sin x . \sin y \)
\( = \frac{4}{5} \times \frac{12}{13} - \frac{3}{5} \times \frac{5}{13} \implies \cos(x + y) = \frac{33}{65} \)
\( \implies x + y = \cos^{-1}\left( \frac{33}{65} \right) \) \( \left[ \because \frac{33}{65} \in [-1, 1] \right] \)
Putting the value of \( x \) and \( y \), we get
\( \cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1}\left( \frac{33}{65} \right) = \text{RHS} \)

Question. Prove that: \( \cos(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}) = \frac{6}{5\sqrt{13}} \).
Answer: Here LHS \( = \cos(\sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2}) \)
Let \( \sin^{-1} \frac{3}{5} = \theta \text{ and } \cot^{-1} \frac{3}{2} = \phi \implies \sin \theta = \frac{3}{5} \text{ and } \cot \phi = \frac{3}{2} \)
\( \implies \cos \theta = \frac{4}{5}, \sin \phi = \frac{2}{\sqrt{13}} \text{ and } \cos \phi = \frac{3}{\sqrt{13}} \)
Now, \( \cos(\theta + \phi) = \cos \theta . \cos \phi - \sin \theta . \sin \phi \)
\( = \frac{4}{5} . \frac{3}{\sqrt{13}} - \frac{3}{5} . \frac{2}{\sqrt{13}} = \frac{12}{5\sqrt{13}} - \frac{6}{5\sqrt{13}} = \frac{6}{5\sqrt{13}} \)

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: LHS \( = \tan\left( \frac{\pi}{4} + \frac{1}{2} \cos^{-1} \frac{a}{b} \right) + \tan\left( \frac{\pi}{4} - \frac{1}{2} \cos^{-1} \frac{a}{b} \right) \)
\( = \tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) \), where \( x = \frac{1}{2} \cos^{-1} \frac{a}{b} \)
\( = \frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x} + \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} \)
\( = \frac{(1 + \tan x)^2 + (1 - \tan x)^2}{1 - \tan^2 x} = \frac{1 + \tan^2 x + 2 \tan x + 1 + \tan^2 x - 2 \tan x}{1 - \tan^2 x} = \frac{2(1 + \tan^2 x)}{1 - \tan^2 x} \)
\( = \frac{2}{\cos 2x} = \frac{2}{\cos(2 . \frac{1}{2} \cos^{-1} \frac{a}{b})} = \frac{2}{\cos(\cos^{-1} \frac{a}{b})} \) \( [\because \cos(\cos^{-1} x) = x \text{ if } x \in [-1, 1]] \)
\( = \frac{2}{a/b} = \frac{2b}{a} = \text{RHS} \) [Here \( \frac{a}{b} \in [-1, 1] \)]

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

Question. If \( \cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha \) prove that \( \frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha \).
Answer: Given, \( \cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha \)
\( \implies \cos^{-1} \left[ \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right] = \alpha \) \( [\because \cos^{-1} x + \cos^{-1} y = \cos^{-1}\{xy - \sqrt{1-x^2}\sqrt{1-y^2}\}] \)
\( \implies \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} = \cos \alpha \implies \frac{xy}{ab} - \sqrt{\frac{b^2 - y^2}{b^2} + \frac{a^2 - x^2}{a^2} + \frac{x^2 y^2}{a^2 b^2}} = \cos \alpha \)
\( \implies \frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2}} \implies \left( \frac{xy}{ab} - \cos \alpha \right)^2 = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \)
\( \implies \frac{x^2 y^2}{a^2 b^2} + \cos^2 \alpha - 2 \frac{xy}{ab} . \cos \alpha = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \)
\( \implies \frac{x^2}{a^2} - 2 \frac{xy}{ab} \cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha \)
\( \implies \frac{x^2}{a^2} - 2 \frac{xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha \). 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: Now, \( \tan^{-1}\left( \frac{\cos x}{1 + \sin x} \right) = \tan^{-1}\left( \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \cos \frac{x}{2} . \sin \frac{x}{2}} \right) \)
\( = \tan^{-1}\left( \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} \right) = \tan^{-1}\left( \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \right) \)
\( = \tan^{-1}\left[ \frac{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}} - \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}} + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}} \right] \) [Divide each term by \( \cos \frac{x}{2} \)]
\( = \tan^{-1}\left[ \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \right] = \tan^{-1}\left[ \frac{\tan \frac{\pi}{4} - \tan \frac{x}{2}}{1 + \tan \frac{\pi}{4} \tan \frac{x}{2}} \right] = \tan^{-1}\left[ \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \right] \)
\( = \frac{\pi}{4} - \frac{x}{2} \) \( \left[ \because \left( \frac{\pi}{4} - \frac{x}{2} \right) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \right] \)
\( \because -\frac{\pi}{2} < x < \frac{\pi}{2} \implies -\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{4} \implies \frac{\pi}{4} > -\frac{x}{2} > -\frac{\pi}{4} \implies \frac{\pi}{4} + \frac{\pi}{4} > \frac{\pi}{4} - \frac{x}{2} > \frac{\pi}{4} - \frac{\pi}{4} \)
\( \implies \frac{\pi}{2} > \frac{\pi}{4} - \frac{x}{2} > 0 \implies \left( \frac{\pi}{4} - \frac{x}{2} \right) \in \left( 0, \frac{\pi}{2} \right) \subset \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)

Question. If \( \tan^{-1} \frac{x-1}{x-2} + \tan^{-1} \frac{x+1}{x+2} = \frac{\pi}{4} \), then find the value of \( x \).
Answer: Given \( \tan^{-1} \frac{x-1}{x-2} + \tan^{-1} \frac{x+1}{x+2} = \frac{\pi}{4} \)
\( \implies \tan^{-1}\left[ \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \times \frac{x+1}{x+2}} \right] = \frac{\pi}{4} \) \( \left[ \text{Using } \tan^{-1} x \pm \tan^{-1} y = \tan^{-1} \frac{x \pm y}{1 \mp xy} \right] \)
\( \implies \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan \frac{\pi}{4} \)
\( \implies \frac{x^2 + x - 2 + x^2 - x - 2}{x^2 - 4 - x^2 + 1} = 1 \implies \frac{2(x^2 - 2)}{-3} = 1 \implies 2x^2 - 4 = -3 \)
\( \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \)
\( \therefore x = \pm \frac{1}{\sqrt{2}} \)

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

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 \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1} \frac{8}{31} \) \( [\because 0 < x < 1 \implies (x+1)(x-1) < 1] \)
\( \implies \tan^{-1} \frac{x+1 + x-1}{1 - (x+1)(x-1)} = \tan^{-1} \frac{8}{31} \)
\( \implies \tan^{-1} \frac{2x}{1 - x^2 + 1} = \tan^{-1} \frac{8}{31} \implies \tan^{-1} \frac{2x}{2 - x^2} = \tan^{-1} \frac{8}{31} \)
\( \implies \frac{2x}{2 - x^2} = \frac{8}{31} \implies 16 - 8x^2 = 62x \)
\( \implies 4x^2 + 31x - 8 = 0 \)
\( \implies 4x^2 + 32x - x - 8 = 0 \implies 4x(x + 8) - 1(x + 8) = 0 \)
\( \implies (x + 8)(4x - 1) = 0 \implies x = -8 \text{ or } x = \frac{1}{4} \)
[\( x = -8 \) is not acceptable]

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

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 \( = \tan^{-1}\left( \frac{1}{2} \right) + \tan^{-1}\left( \frac{1}{5} \right) + \tan^{-1}\left( \frac{1}{8} \right) \)
\( = \tan^{-1} \left[ \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \times \frac{1}{5}} \right] + \tan^{-1}\left( \frac{1}{8} \right) \) \( \left[ \because \frac{1}{2} \times \frac{1}{5} = \frac{1}{10} < 1 \right] \)
\( = \tan^{-1} \left( \frac{7}{9} \right) + \tan^{-1}\left( \frac{1}{8} \right) = \tan^{-1} \left[ \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \times \frac{1}{8}} \right] = \tan^{-1}\left( \frac{65/72}{65/72} \right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{RHS} \)

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 \( = \sin^{-1}\left( \frac{4}{5} \right) + \tan^{-1}\left( \frac{5}{12} \right) \)
\( = \sin^{-1}\left( \frac{4}{5} \right) + \sin^{-1} \left[ \frac{\frac{5}{12}}{\sqrt{1 + \left( \frac{5}{12} \right)^2}} \right] \) \( \left[ \because \tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right) \right] \)
\( = \sin^{-1}\left( \frac{4}{5} \right) + \sin^{-1} \left[ \frac{5/12}{13/12} \right] = \sin^{-1}\left( \frac{4}{5} \right) + \sin^{-1}\left( \frac{5}{13} \right) \)
\( = \sin^{-1} \left[ \frac{4}{5} \sqrt{1 - \left( \frac{5}{13} \right)^2} + \frac{5}{13} \sqrt{1 - \left( \frac{4}{5} \right)^2} \right] \)
\( = \sin^{-1} \left[ \frac{4}{5} \sqrt{\left( \frac{12}{13} \right)^2} + \frac{5}{13} \sqrt{\left( \frac{3}{5} \right)^2} \right] \)
\( = \sin^{-1} \left[ \frac{4}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{3}{5} \right] = \sin^{-1}\left( \frac{48}{65} + \frac{15}{65} \right) = \sin^{-1}\left( \frac{63}{65} \right) = \text{RHS} \) 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: \( \tan^{-1}\left( \frac{x-2}{x-3} \right) + \tan^{-1}\left( \frac{x+2}{x+3} \right) = \frac{\pi}{4}, |x| < 1 \)
\( \implies \tan^{-1} \left[ \frac{\frac{x-2}{x-3} + \frac{x+2}{x+3}}{1 - \left( \frac{x-2}{x-3} \right) \left( \frac{x+2}{x+3} \right)} \right] = \frac{\pi}{4} \) \( \left[ \because \frac{x-2}{x-3} . \frac{x+2}{x+3} = \frac{x^2-4}{x^2-9} < 1 \text{ for } |x| < 1 \right] \)
\( \implies \tan^{-1} \left[ \frac{(x-2)(x+3) + (x+2)(x-3)}{(x-3)(x+3) - (x-2)(x+2)} \right] = \frac{\pi}{4} \)
\( \implies \tan^{-1} \left[ \frac{x^2+3x-2x-6 + x^2-3x+2x-6}{x^2-9-x^2+4} \right] = \frac{\pi}{4} \)
\( \implies \tan^{-1} \left[ \frac{2x^2-12}{-5} \right] = \frac{\pi}{4} \implies \frac{2x^2-12}{-5} = \tan \frac{\pi}{4} \implies \frac{2x^2-12}{-5} = 1 \)
\( \implies 2x^2 - 12 = -5 \implies 2x^2 = 7 \implies x^2 = \frac{7}{2} \)
\( \implies x = \pm \sqrt{\frac{7}{2}} \), 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, \( (\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8} \)
\( \implies (\tan^{-1} x)^2 + \left( \frac{\pi}{2} - \tan^{-1} x \right)^2 = \frac{5\pi^2}{8} \)
\( \implies (\tan^{-1} x)^2 + (\tan^{-1} x)^2 + \frac{\pi^2}{4} - \pi \tan^{-1} x = \frac{5\pi^2}{8} \)
\( \implies 2(\tan^{-1} x)^2 - \pi \tan^{-1} x + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0 \)
\( \implies 2(\tan^{-1} x)^2 - \pi \tan^{-1} x - \frac{3\pi^2}{8} = 0 \) ...(i)
Let \( \tan^{-1} x = y \), then (i) becomes
\( 2y^2 - \pi y - \frac{3\pi^2}{8} = 0 \implies 16y^2 - 8\pi y - 3\pi^2 = 0 \)
\( \implies 16y^2 - 12\pi y + 4\pi y - 3\pi^2 = 0 \implies 4y(4y - 3\pi) + \pi(4y - 3\pi) = 0 \)
\( \implies (4y - 3\pi)(4y + \pi) = 0 \implies y = -\frac{\pi}{4} \text{ or } y = \frac{3\pi}{4} \)
\( \implies \tan^{-1} x = -\frac{\pi}{4} \) \( \left[ \because \frac{3\pi}{4} \text{ does not belongs to domain of } \tan^{-1} x \text{ i.e., } \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \right] \)
\( \implies x = \tan\left( -\frac{\pi}{4} \right) = -1 \)

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 \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \implies \tan^{-1} x + \tan^{-1} y = \frac{\pi}{2} - \tan^{-1} z \)
\( \implies \tan^{-1} x + \tan^{-1} y = \cot^{-1} z \implies \tan^{-1} \left( \frac{x+y}{1-xy} \right) = \tan^{-1} \frac{1}{z} \)
\( \implies \frac{x+y}{1-xy} = \frac{1}{z} \implies xz + yz = 1 - xy \implies xy + yz + zx = 1 \)

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

Question. Find the value of \( \cot\left\{ \frac{1}{2} \cos^{-1} \frac{2x}{1+x^2} + \frac{1}{2} \sin^{-1} \frac{1-y^2}{1+y^2} \right\} |x| < 1, y > 0 \text{ and } xy < 1 \).
Answer: \( \cot\left[ \frac{1}{2} \cos^{-1} \frac{2x}{1+x^2} + \frac{1}{2} \sin^{-1} \frac{1-y^2}{1+y^2} \right] = \cot\left[ \frac{1}{2} \left( \frac{\pi}{2} - \sin^{-1} \frac{2x}{1+x^2} \right) + \frac{1}{2} \left( \frac{\pi}{2} - \cos^{-1} \frac{1-y^2}{1+y^2} \right) \right] \)
\( = \cot \left[ \frac{\pi}{4} - \frac{1}{2} \sin^{-1} \frac{2x}{1+x^2} + \frac{\pi}{4} - \frac{1}{2} \cos^{-1} \frac{1-y^2}{1+y^2} \right] \)
\( = \cot \left[ \frac{\pi}{2} - 2 \tan^{-1} x - 2 \tan^{-1} y \right] \)
\( = \cot \left[ \frac{\pi}{2} - (\tan^{-1} x + \tan^{-1} y) \right] = \tan(\tan^{-1} x + \tan^{-1} y) \)
\( = \tan \left( \tan^{-1} \left( \frac{x+y}{1-xy} \right) \right) = \frac{x+y}{1-xy} \) \( [\because xy < 1] \)

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: \( \tan^{-1}\left( \frac{x+1}{x-1} \right) + \tan^{-1}\left( \frac{x-1}{x} \right) = -\tan^{-1} 7 \)
\( \implies \tan^{-1} \left[ \frac{\left( \frac{x+1}{x-1} \right) + \left( \frac{x-1}{x} \right)}{1 - \left( \frac{x+1}{x-1} \right) \left( \frac{x-1}{x} \right)} \right] = -\tan^{-1} 7, \text{ if } \left( \frac{x+1}{x-1} \right)\left( \frac{x-1}{x} \right) < 1 \) ...(i)
\( \implies \tan^{-1} \left[ \frac{x(x+1) + (x-1)^2}{(x-1)x - (x+1)(x-1)} \right] = \tan^{-1}(-7) \)
\( \implies \frac{(x^2+x) + (x^2+1-2x)}{(x^2-x) - (x^2-1)} = \tan[-\tan^{-1} 7] = \tan(\tan^{-1}(-7)) = -7 \)
\( \implies \frac{2x^2-x+1}{-x+1} = -7 \implies 2x^2-8x+8=0 \implies 2(x^2-4x+4)=0 \)
\( \implies (x-2)^2 = 0 \implies x = 2 \)
Let us now verify whether \( x = 2 \) satisfies the equation (i)
For \( x = 2, \left( \frac{x+1}{x-1} \right) \left( \frac{x-1}{x} \right) = 3 \times \frac{1}{2} = \frac{3}{2} \), which is not less than 1
Hence, this value does not satisfy the equation (i)
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 \( \sin^{-1}\sqrt{x^2+x+1} = \theta \)
\( \implies \sin \theta = \frac{\sqrt{x^2+x+1}}{1} \implies \tan \theta = \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} \) \( \left[ \because \tan \theta = \frac{\sin \theta}{\cos \theta} \right] \)
\( \therefore \theta = \tan^{-1} \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} = \sin^{-1} \sqrt{x^2+x+1} \)
On putting the value of \( \theta \) in equation (i), we get
\( \tan^{-1}\sqrt{x(x+1)} + \tan^{-1} \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}} = \frac{\pi}{2} \)
We know that, \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left( \frac{x+y}{1-xy} \right), xy < 1 \)
\( \therefore \tan^{-1} \left[ \frac{\sqrt{x(x+1)} + \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}}{1 - \sqrt{x(x+1)} . \frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}} \right] = \frac{\pi}{2} \)
\( \implies \tan^{-1} \left[ \frac{\sqrt{x^2+x} + \frac{\sqrt{x^2+x+1}}{\sqrt{-(x^2+x)}}}{1 - \sqrt{(x^2+x)} . \frac{\sqrt{x^2+x+1}}{\sqrt{-(x^2+x)}}} \right] = \frac{\pi}{2} \)
\( \implies \frac{x^2+x+\sqrt{-(x^2+x+1)}}{[1 - \sqrt{-(x^2+1)}]\sqrt{x^2+x}} = \tan \frac{\pi}{2} = \frac{1}{0} \)
\( \implies [1 - \sqrt{-(x^2+x+1)}]\sqrt{x^2+x} = 0 \)
\( \implies -(x^2+x+1) = 1 \text{ or } x^2+x = 0 \)
\( \implies -x^2-x-1 = 1 \text{ or } x(x+1) = 0 \)
\( \implies x^2+x+2 = 0 \text{ or } x(x+1) = 0 \)
\( \therefore x = \frac{-1 \pm \sqrt{1-4 \times 2}}{2} = \frac{-1 \pm \sqrt{-7}}{2} \), which is not real or \( x = 0 \) or \( x = -1 \)
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 \( \sin^{-1} x = A \implies \sin A = x \)
\( \sin^{-1} y = B \implies \sin B = y \)
\( \sin^{-1} z = C \implies \sin C = z \)
Given, \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \)
\( \implies A + B + C = \pi \implies 2A + 2B + 2C = 2\pi \)
\( \therefore \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \) [Using trigonometric property]
\( \implies 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C = 4 \sin A \sin B \sin C \)
\( \implies 2 \sin A . \sqrt{1 - \sin^2 A} + 2 \sin B . \sqrt{1 - \sin^2 B} + 2 \sin C . \sqrt{1 - \sin^2 C} = 4 \sin A \sin B \sin C \)
\( \implies 2x\sqrt{1-x^2} + 2y\sqrt{1-y^2} + 2z\sqrt{1-z^2} = 4xyz \)
\( \implies x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz \) Hence proved.

Question. If \( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \), then prove that \( a + b + c = abc \).
Answer: Firstly, let us assume
\( \tan^{-1} a = \alpha \implies \tan \alpha = a \)
\( \tan^{-1} b = \beta \implies \tan \beta = b \)
\( \tan^{-1} c = \gamma \implies \tan \gamma = c \)
Now, given that
\( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \implies \alpha + \beta + \gamma = \pi \)
\( \therefore \alpha + \beta = \pi - \gamma \)
Taking tangent on both sides, we have
\( \tan(\alpha + \beta) = \tan(\pi - \gamma) \)
\( \implies \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = -\tan \gamma \)
\( \implies \tan \alpha + \tan \beta = -\tan \gamma(1 - \tan \alpha \tan \beta) \)
\( \implies \tan \alpha + \tan \beta = -\tan \gamma + \tan \alpha \tan \beta \tan \gamma \)
\( \implies \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \)
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 \( = 2 \tan^{-1}\left\{ \tan \frac{\alpha}{2} \tan\left( \frac{\pi}{4} - \frac{\beta}{2} \right) \right\} \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} . \tan\left( \frac{\pi}{4} - \frac{\beta}{2} \right)}{1 - \tan^2 \frac{\alpha}{2} . \tan^2\left( \frac{\pi}{4} - \frac{\beta}{2} \right)} \right] \) \( \left[ \because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} . \left( \frac{1 - \tan \frac{\beta}{2}}{1 + \tan \frac{\beta}{2}} \right)}{1 - \tan^2 \frac{\alpha}{2} . \left( \frac{1 - \tan \frac{\beta}{2}}{1 + \tan \frac{\beta}{2}} \right)^2} \right] \) \( \left[ \because \tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} \left( 1 - \tan \frac{\beta}{2} \right) \left( 1 + \tan \frac{\beta}{2} \right)}{(1 + \tan \frac{\beta}{2})^2 - \tan^2 \frac{\alpha}{2} (1 - \tan \frac{\beta}{2})^2} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} (1 - \tan^2 \frac{\beta}{2})}{(1 + \tan^2 \frac{\beta}{2} + 2 \tan \frac{\beta}{2}) - \tan^2 \frac{\alpha}{2} (1 + \tan^2 \frac{\beta}{2} - 2 \tan \frac{\beta}{2})} \right] \)
\( = \tan^{-1} \left[ \frac{2 \tan \frac{\alpha}{2} (1 - \tan^2 \frac{\beta}{2})}{(1 + \tan^2 \frac{\beta}{2}) (1 - \tan^2 \frac{\alpha}{2}) + 2 \tan \frac{\beta}{2} (1 + \tan^2 \frac{\alpha}{2})} \right] \)
\( = \tan^{-1} \left[ \frac{\frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} \cdot \frac{1 - \tan^2 \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}}}{\frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} + \frac{2 \tan \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}}} \right] \) [Dividing N' and D' by \( (1 + \tan^2 \frac{\alpha}{2})(1 + \tan^2 \frac{\beta}{2}) \)]
\( = \tan^{-1} \left[ \frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta} \right] = \text{RHS} \)

Question. Solve the equation \( \tan^{-1}\sqrt{x^2+x} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2} \).
Answer: Given equation exists, if
\( x^2+x \ge 0 \) and \( 0 < \sqrt{x^2+x+1} \le 1 \) \( [\because x^2+x+1 \text{ is always greater than zero}] \)
Now, \( x^2+x \ge 0 \) and \( x^2+x+1 \le 1 \)
\( \implies x^2+x \ge 0 \) and \( x^2+x \le 0 \)
\( \implies x^2+x = 0 \) i.e., \( x(x+1) = 0 \)
Hence, \( x = 0 \) and \( -1 \) are the solutions of the given equation.

Question. If \( a_1, a_2, a_3, ..., a_n \) is an arithmetic progression with common difference \( d \), then evaluate the following expression. \( \tan[\tan^{-1}(\frac{d}{1+a_1a_2}) + \tan^{-1}(\frac{d}{1+a_2a_3}) + \tan^{-1}(\frac{d}{1+a_3a_4}) + ... + \tan^{-1}(\frac{d}{1+a_{n-1}a_n})] \).
Answer: We have, \( a_1 = a, a_2 = a + d, a_3 = a + 2d \)
and \( d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = ... = a_n - a_{n-1} \)
Given that, \( \tan\left[ \tan^{-1}\left( \frac{d}{1+a_1a_2} \right) + \tan^{-1}\left( \frac{d}{1+a_2a_3} \right) + \tan^{-1}\left( \frac{d}{1+a_3a_4} \right) + ... + \tan^{-1}\left( \frac{d}{1+a_{n-1}a_n} \right) \right] \)
\( = \tan\left[ \tan^{-1}\left( \frac{a_2-a_1}{1+a_1a_2} \right) + \tan^{-1}\left( \frac{a_3-a_2}{1+a_2a_3} \right) + ... + \tan^{-1}\left( \frac{a_n-a_{n-1}}{1+a_{n-1}a_n} \right) \right] \)
\( = \tan[(\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + ... + (\tan^{-1} a_n - \tan^{-1} a_{n-1})] \)
\( = \tan[\tan^{-1} a_n - \tan^{-1} a_1] \)
\( = \tan \left[ \tan^{-1} \left( \frac{a_n-a_1}{1+a_n a_1} \right) \right] \) \( [\because \tan^{-1} x - \tan^{-1} y = \tan^{-1}\left( \frac{x-y}{1+xy} \right)] \)
\( = \frac{a_n-a_1}{1+a_n a_1} \) \( [\because \tan(\tan^{-1} x) = x] \)