CBSE Class 12 Mathematics HOTs Inverse Trigonometric Functions Set 03

Question. Solve for \( x \): \( \cos(\tan^{-1}x) = \sin(\cot^{-1}\frac{3}{4}) \)
Answer: We have, \( \cos(\tan^{-1}x) = \sin(\cot^{-1}\frac{3}{4}) \)
Let \( \tan^{-1}x = \theta \Rightarrow \tan\theta = x \)
\( \Rightarrow \cos\theta = \frac{1}{\sqrt{1+x^2}} \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \)
Also, let \( \cot^{-1}\frac{3}{4} = \beta \Rightarrow \cot\beta = \frac{3}{4} \)
\( \Rightarrow \sin\beta = \frac{4}{5} \Rightarrow \beta = \sin^{-1}\frac{4}{5} \)
So, \( \cos(\tan^{-1}x) = \sin(\cot^{-1}\frac{3}{4}) \)
\( \Rightarrow \cos\left[\cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right] = \sin\left[\sin^{-1}\left(\frac{4}{5}\right)\right] \)
\( \Rightarrow \frac{1}{\sqrt{1+x^2}} = \frac{4}{5} \)
\( \Rightarrow 16 + 16x^2 = 25 \Rightarrow 16x^2 = 9 \Rightarrow x = \pm 3/4 \).
Hence, the values of \( x \) are \( 3/4, -3/4 \).

Question. Prove that: \( \cot^{-1} 7 + \dots + \cot^{-1} 8 + \dots + \cot^{-1} 18 = \cot^{-1} 3 \).
Answer: L.H.S. \( = \cot^{-1} 7 + \cot^{-1} 8 + \cot^{-1} 18 \)
\( = \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} + \tan^{-1}\frac{1}{18} \)
\( = \tan^{-1}\left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}} \right) + \tan^{-1}\frac{1}{18} \)
\( = \tan^{-1}\frac{3}{11} + \tan^{-1}\frac{1}{18} \)
\( = \tan^{-1}\left( \frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \cdot \frac{1}{18}} \right) = \tan^{-1}\left( \frac{\frac{54 + 11}{198}}{\frac{198 - 3}{198}} \right) = \tan^{-1}\left(\frac{65}{195}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
\( = \cot^{-1} 3 = \text{R.H.S.} \)
Hence proved.

Question. Prove that: \( \cos^{-1}(x) + \cos^{-1}\left[\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right] = \frac{\pi}{3} \)
Answer: L.H.S. \( = \cos^{-1}x + \cos^{-1}\left[ \frac{x}{2} + \frac{\sqrt{3-3x^2}}{2} \right] \)
\( = \cos^{-1}x + \cos^{-1}\left[ x \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \sqrt{1-x^2} \right] \)
\( = \cos^{-1}x + \cos^{-1}\left[ x \cdot \frac{1}{2} + \sqrt{1 - \left(\frac{1}{2}\right)^2} \cdot \sqrt{1-x^2} \right] \)
\( = \cos^{-1}x + \cos^{-1}\left(\frac{1}{2}\right) - \cos^{-1}x \)
\( = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1} x + 2\cot^{-1} x = \frac{2\pi}{3} \)
Answer: \( \tan^{-1} x + 2\cot^{-1} x = \frac{2\pi}{3} \)
\( \Rightarrow \frac{\pi}{2} - \cot^{-1} x + 2\cot^{-1} x = \frac{2\pi}{3} \)
\( \Rightarrow \cot^{-1} x = \frac{2\pi}{3} - \frac{\pi}{2} \)
\( \Rightarrow \cot^{-1} x = \frac{4\pi - 3\pi}{6} \)
\( \Rightarrow \cot^{-1} x = \frac{\pi}{6} \Rightarrow x = \cot\frac{\pi}{6} \Rightarrow x = \sqrt{3} \)

Question. Prove that: \( \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \cos^{-1}\frac{36}{85} \)
Answer: L.H.S. \( = \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} \)
\( = \sin^{-1}\left( \frac{3}{5}\sqrt{1 - \left(\frac{8}{17}\right)^2} + \frac{8}{17}\sqrt{1 - \left(\frac{3}{5}\right)^2} \right) \)
\( = \sin^{-1}\left( \frac{3}{5} \cdot \frac{15}{17} + \frac{8}{17} \cdot \frac{4}{5} \right) \)
\( = \sin^{-1}\left( \frac{45}{85} + \frac{32}{85} \right) = \sin^{-1}\left(\frac{77}{85}\right) \)
\( = \cos^{-1}\left(\sqrt{1 - \left(\frac{77}{85}\right)^2}\right) = \cos^{-1}\left(\sqrt{\frac{7225 - 5929}{7225}}\right) \)
\( = \cos^{-1}\left(\sqrt{\frac{1296}{7225}}\right) = \cos^{-1}\left(\frac{36}{85}\right) = \text{R.H.S.} \)

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

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: L.H.S. \( = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) \)
\( = \tan^{-1}\left( \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}} \right) + \tan^{-1}\left(\frac{1}{8}\right) \)
\( = \tan^{-1}\left( \frac{7/10}{9/10} \right) + \tan^{-1}\left(\frac{1}{8}\right) \)
\( = \tan^{-1}\left(\frac{7}{9}\right) + \tan^{-1}\left(\frac{1}{8}\right) \)
\( = \tan^{-1}\left( \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \cdot \frac{1}{8}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{56 + 9}{72}}{\frac{72 - 7}{72}} \right) = \tan^{-1}\left(\frac{65}{65}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.} \)

Question. Show that: \( \tan\left( \frac{1}{2}\sin^{-1}\frac{3}{4} \right) = \frac{4-\sqrt{7}}{3} \)
Answer: Put \( \sin^{-1}\frac{3}{4} = \theta \Rightarrow \sin\theta = \frac{3}{4} \)
\( \Rightarrow \cos\theta = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \frac{\sqrt{7}}{4} \)
Now, \( \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) = \tan\frac{\theta}{2} \)
\( = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} = \sqrt{\frac{1 - \frac{\sqrt{7}}{4}}{1 + \frac{\sqrt{7}}{4}}} = \sqrt{\frac{4 - \sqrt{7}}{4 + \sqrt{7}}} = \sqrt{\frac{(4 - \sqrt{7})^2}{16 - 7}} = \frac{4-\sqrt{7}}{\sqrt{9}} = \frac{4-\sqrt{7}}{3} \)

Question. Write the value of the following: \( \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right) \)
Answer: \( \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right) \)
\( = \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{\frac{a}{b}-1}{1+\frac{a}{b}}\right) \)
\( = \tan^{-1}\left(\frac{a}{b}\right) - \left[ \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}(1) \right] \)
\( = \tan^{-1}(1) = \frac{\pi}{4} \)

Question. Prove that: \( \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36} \)
Answer: Let \( \sin^{-1}\frac{8}{17} = x \Rightarrow \sin x = \frac{8}{17} \Rightarrow \tan x = \frac{8}{15} \)
Let \( \sin^{-1}\frac{3}{5} = y \Rightarrow \sin y = \frac{3}{5} \Rightarrow \tan y = \frac{3}{4} \)
L.H.S. \( = \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{8}{15} + \tan^{-1}\frac{3}{4} \)
\( = \tan^{-1}\left( \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{32 + 45}{60}}{\frac{60 - 24}{60}} \right) = \tan^{-1}\left(\frac{77}{36}\right) = \text{R.H.S.} \)

Question. Solve for \( x \): \( \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} \)
Answer: We have, \( \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} \quad \dots(1) \)
\( \Rightarrow \sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1}x \)
\( \Rightarrow 1-x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right) = \cos(2\sin^{-1}x) \)
Let \( \theta = \sin^{-1}x \Rightarrow x = \sin\theta \)
\( \therefore 1-x = 1 - 2\sin^2\theta \Rightarrow 1-x = 1 - 2x^2 \)
\( \Rightarrow 2x^2 - x = 0 \Rightarrow x(2x - 1) = 0 \Rightarrow x = 0 \text{ or } x = \frac{1}{2} \)
For \( x = \frac{1}{2} \),
L.H.S. of (1) \( = \sin^{-1}\left(1 - \frac{1}{2}\right) - 2\sin^{-1}\frac{1}{2} \)
\( = \sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = -\sin^{-1}\frac{1}{2} = -\frac{\pi}{6} \neq \frac{\pi}{2} \)
\( \therefore x = \frac{1}{2} \) is not a solution of (1).
Hence, \( x = 0 \) is the only solution of (1).

Question. Prove that \( \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) = \frac{\pi}{4} - \frac{x}{2} \text{, } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
Answer: L.H.S. \( = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
\( = \tan^{-1}\left( \frac{\sin\left(\frac{\pi}{2}-x\right)}{1 + \cos\left(\frac{\pi}{2}-x\right)} \right) \)
\( = \tan^{-1}\left( \frac{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \right) \)
\( = \tan^{-1}\left( \tan\left(\frac{\pi}{4}-\frac{x}{2}\right) \right) = \frac{\pi}{4} - \frac{x}{2} = \text{R.H.S.} \)

Question. Prove the following: \( \cos\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right) = \frac{6}{5\sqrt{13}} \)
Answer: Let \( \sin^{-1}\frac{3}{5} = \theta \Rightarrow \sin\theta = \frac{3}{5} \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \theta = \tan^{-1}\frac{3}{4} \)
And \( \cot^{-1}\frac{3}{2} = \phi \Rightarrow \cot\phi = \frac{3}{2} \Rightarrow \tan\phi = \frac{2}{3} \Rightarrow \phi = \tan^{-1}\frac{2}{3} \)
Thus, \( \sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2} = \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{2}{3} \)
\( = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4}\cdot\frac{2}{3}} \right) = \tan^{-1}\left( \frac{17/12}{6/12} \right) = \tan^{-1}\left(\frac{17}{6}\right) \)
Let \( \tan^{-1}\left(\frac{17}{6}\right) = \alpha \Rightarrow \tan\alpha = \frac{17}{6} \Rightarrow \cos\alpha = \frac{6}{\sqrt{6^2 + 17^2}} = \frac{6}{\sqrt{36 + 289}} = \frac{6}{5\sqrt{13}} \)
Now, L.H.S. \( = \cos\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right) = \cos\alpha = \frac{6}{5\sqrt{13}} = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \)
Answer: We have, \( \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}\left( \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)} \right) = \frac{\pi}{4} \)
\( \Rightarrow \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan\frac{\pi}{4} = 1 \)
\( \Rightarrow \frac{(x^2+x-2) + (x^2-x-2)}{(x^2-4) - (x^2-1)} = 1 \)
\( \Rightarrow \frac{2x^2 - 4}{-3} = 1 \)
\( \Rightarrow 2x^2 - 4 = -3 \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm\frac{1}{\sqrt{2}} \)

Question. Prove that: \( \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right) - \tan^{-1}\left(\frac{8}{19}\right) = \frac{\pi}{4} \)
Answer: L.H.S. \( = \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{3}{5}\right) - \tan^{-1}\left(\frac{8}{19}\right) \)
\( = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4}\cdot\frac{3}{5}} \right) - \tan^{-1}\left(\frac{8}{19}\right) \)
\( = \tan^{-1}\left(\frac{27/20}{11/20}\right) - \tan^{-1}\left(\frac{8}{19}\right) \)
\( = \tan^{-1}\left(\frac{27}{11}\right) - \tan^{-1}\left(\frac{8}{19}\right) \)
\( = \tan^{-1}\left( \frac{\frac{27}{11} - \frac{8}{19}}{1 + \frac{27}{11}\cdot\frac{8}{19}} \right) = \tan^{-1}\left( \frac{\frac{513 - 88}{209}}{\frac{209 + 216}{209}} \right) \)
\( = \tan^{-1}\left(\frac{425}{425}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.} \)

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

Question. Prove that: \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \)
Answer: L.H.S. \( = 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} \right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{1}{3/4}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3}\cdot\frac{1}{7}} \right) = \tan^{-1}\left( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right) = \tan^{-1}\left(\frac{31}{17}\right) = \text{R.H.S.} \)

Question. Prove that: \( 2\tan^{-1}\frac{3}{4} - \tan^{-1}\frac{17}{31} = \frac{\pi}{4} \)
Answer: L.H.S. \( = 2\tan^{-1}\frac{3}{4} - \tan^{-1}\frac{17}{31} \)
\( = \tan^{-1}\left( \frac{2 \cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} \right) - \tan^{-1}\frac{17}{31} \)
\( = \tan^{-1}\left( \frac{3/2}{7/16} \right) - \tan^{-1}\frac{17}{31} \)
\( = \tan^{-1}\left(\frac{24}{7}\right) - \tan^{-1}\frac{17}{31} \)
\( = \tan^{-1}\left( \frac{\frac{24}{7} - \frac{17}{31}}{1 + \frac{24}{7}\cdot\frac{17}{31}} \right) = \tan^{-1}\left( \frac{24 \cdot 31 - 17 \cdot 7}{7 \cdot 31 + 24 \cdot 17} \right) \)
\( = \tan^{-1}\left( \frac{744 - 119}{217 + 408} \right) = \tan^{-1}\left(\frac{625}{625}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3} \text{, } -1 < x < 1 \)
Answer: We have, \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3} \)
\( \because \cot^{-1}z = \tan^{-1}\left(\frac{1}{z}\right) \) for \( z > 0 \)
\( \Rightarrow \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \)
\( \Rightarrow 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \)
\( \Rightarrow \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{6} \)
\( \Rightarrow \frac{2x}{1-x^2} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow 2\sqrt{3}x = 1 - x^2 \Rightarrow x^2 + 2\sqrt{3}x - 1 = 0 \)
\( \Rightarrow x = \frac{-2\sqrt{3} \pm \sqrt{12 + 4}}{2} = -\sqrt{3} \pm 2 \)
Since \( -1 < x < 1 \), we reject \( -\sqrt{3}-2 \) as it is less than \(-1\).
\( \therefore x = 2 - \sqrt{3} \)

Question. Prove that: \( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \frac{1}{2}\tan^{-1}\frac{4}{3} \)
Answer: To prove: \( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \frac{1}{2}\tan^{-1}\frac{4}{3} \)
\( \Rightarrow 2\left[ \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} \right] = \tan^{-1}\frac{4}{3} \)
Now L.H.S. \( = 2\left[ \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} \right] \)
\( = 2 \tan^{-1}\left( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4}\cdot\frac{2}{9}} \right) = 2 \tan^{-1}\left( \frac{17/36}{34/36} \right) = 2\tan^{-1}\frac{1}{2} \)
\( = \tan^{-1}\left( \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} \right) = \tan^{-1}\left(\frac{1}{3/4}\right) = \tan^{-1}\frac{4}{3} = \text{R.H.S.} \)

Question. Solve for \( x \): \( \cos(2\sin^{-1}x) = \frac{1}{9} \text{, } x > 0 \)
Answer: The given equation is \( \cos(2\sin^{-1}x) = \frac{1}{9} \quad (x > 0) \quad \dots(1) \)
Put \( \sin^{-1}x = \theta \Rightarrow x = \sin\theta \)
\( \therefore \) Eq. (1) \( \Rightarrow \cos 2\theta = \frac{1}{9} \Rightarrow 1 - 2\sin^2\theta = \frac{1}{9} \)
\( \Rightarrow 2\sin^2\theta = 1 - \frac{1}{9} = \frac{8}{9} \)
\( \Rightarrow x^2 = \frac{4}{9} \Rightarrow x = \frac{2}{3} \) (\( \because x > 0 \))

Question. Prove that: \( \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) \text{, } x \in (0,1) \)
Answer: Putting \( x = \tan^2\theta \), we get
R.H.S. \( = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \)
\( = \frac{1}{2}\cos^{-1}(\cos 2\theta) = \frac{1}{2}(2\theta) = \theta \)
\( = \tan^{-1}\sqrt{x} = \text{L.H.S.} \)
\( [ \because x = \tan^2\theta \Rightarrow \tan\theta = \sqrt{x} \Rightarrow \theta = \tan^{-1}\sqrt{x} ] \)
\( \therefore \tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right) \)

Question. Prove that: \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \)
Answer: Let \( \text{L.H.S.} = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \)
We know that for \( x > 0, y > 0 \) and \( xy > 1 \), \( \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).
Here \( 2 \cdot 3 = 6 > 1 \).
So, \( \tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}\left(\frac{2+3}{1-2\cdot 3}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) = \pi + \tan^{-1}(-1) = \pi - \tan^{-1}(1) \).
Thus, L.H.S. \( = \tan^{-1}(1) + \pi - \tan^{-1}(1) = \pi = \text{R.H.S.} \)

Question. Prove that: \( \cos\left[\tan^{-1}\left\{\sin\left(\cot^{-1}x\right)\right\}\right] = \sqrt{\frac{1+x^2}{2+x^2}} \)
Answer: L.H.S. \( = \cos\left[\tan^{-1}\left\{\sin\left(\cot^{-1}x\right)\right\}\right] \)
Let \( \cot^{-1}x = \theta \Rightarrow x = \cot\theta \Rightarrow x^2 = \cot^2\theta \)
\( \Rightarrow \csc^2\theta - 1 = x^2 \Rightarrow \csc^2\theta = 1 + x^2 \)
\( \Rightarrow \csc\theta = \sqrt{1 + x^2} \Rightarrow \sin\theta = \frac{1}{\sqrt{1 + x^2}} \)
Now, L.H.S. \( = \cos\left[ \tan^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \right] \)
Let \( \tan^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) = \phi \Rightarrow \tan\phi = \frac{1}{\sqrt{1+x^2}} \)
\( \Rightarrow \sec^2\phi - 1 = \frac{1}{1+x^2} \Rightarrow \sec^2\phi = 1 + \frac{1}{1+x^2} = \frac{2+x^2}{1+x^2} \)
\( \Rightarrow \cos^2\phi = \frac{1+x^2}{2+x^2} \Rightarrow \cos\phi = \sqrt{\frac{1+x^2}{2+x^2}} \)
\( \therefore \text{L.H.S.} = \cos\phi = \sqrt{\frac{1+x^2}{2+x^2}} = \text{R.H.S.} \)

Question. Prove that: \( \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) \)
Answer: L.H.S. \( = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right) \)
\( = \tan^{-1}\left( \frac{x + \frac{2x}{1-x^2}}{1 - x\left(\frac{2x}{1-x^2}\right)} \right) \)
\( = \tan^{-1}\left( \frac{\frac{x(1-x^2) + 2x}{1-x^2}}{\frac{1-x^2 - 2x^2}{1-x^2}} \right) \)
\( = \tan^{-1}\left( \frac{x - x^3 + 2x}{1 - 3x^2} \right) = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1}\frac{x}{2} + \tan^{-1}\frac{x}{3} = \frac{\pi}{4} \text{, } 0 < x < \sqrt{6} \)
Answer: \( \tan^{-1}\frac{x}{2} + \tan^{-1}\frac{x}{3} = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}\left( \frac{\frac{x}{2} + \frac{x}{3}}{1 - \frac{x}{2}\cdot\frac{x}{3}} \right) = \frac{\pi}{4} \) (for \( \frac{x}{2}\cdot\frac{x}{3} < 1 \))
\( \Rightarrow \tan^{-1}\left( \frac{\frac{5x}{6}}{\frac{6 - x^2}{6}} \right) = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}\left(\frac{5x}{6 - x^2}\right) = \frac{\pi}{4} \)
\( \Rightarrow \frac{5x}{6-x^2} = \tan\frac{\pi}{4} = 1 \Rightarrow x^2 + 5x - 6 = 0 \)
\( \Rightarrow (x+6)(x-1) = 0 \Rightarrow x = -6, x = 1 \)
But \( x = -6 \) does not satisfy the given condition \( 0 < x < \sqrt{6} \).
\( \therefore x = 1 \) is the only solution.

Question. Solve for \( x \): \( \tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}\left(\frac{8}{79}\right) \text{, } x > 0 \)
Answer: We have, \( \tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}\left(\frac{8}{79}\right) \)
\( \Rightarrow \tan^{-1}\left( \frac{(x+2) + (x-2)}{1 - (x+2)(x-2)} \right) = \tan^{-1}\left(\frac{8}{79}\right) \)
\( \Rightarrow \frac{2x}{1 - (x^2 - 4)} = \frac{8}{79} \)
\( \Rightarrow \frac{2x}{5-x^2} = \frac{8}{79} \Rightarrow 79x = 4(5-x^2) \)
\( \Rightarrow 4x^2 + 79x - 20 = 0 \)
\( \Rightarrow (4x-1)(x+20) = 0 \)
\( \Rightarrow x = \frac{1}{4} \) or \( x = -20 \)
Since \( x > 0 \), we have \( x = \frac{1}{4} \) as the only solution.

Question. Prove that: \( 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4} \)
Answer: L.H.S. \( = 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} \)
\( = \tan^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) + \tan^{-1}\frac{1}{7} \)
\( = \tan^{-1}\left(\frac{2/3}{8/9}\right) + \tan^{-1}\frac{1}{7} = \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\frac{1}{7} \)
\( = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4}\cdot\frac{1}{7}}\right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.} \)

Question. Solve for \( x \): \( \cos^{-1}x + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \)
Answer: We have \( \cos^{-1}x + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \)
\( \Rightarrow \cos^{-1}x = \frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right) \)
\( \Rightarrow x = \cos\left( \frac{\pi}{6} - \sin^{-1}\left(\frac{x}{2}\right) \right) \)
\( = \cos\frac{\pi}{6}\cos\left(\sin^{-1}\frac{x}{2}\right) + \sin\frac{\pi}{6}\sin\left(\sin^{-1}\frac{x}{2}\right) \)
\( \Rightarrow x = \frac{\sqrt{3}}{2} \sqrt{1 - \frac{x^2}{4}} + \frac{1}{2} \cdot \frac{x}{2} \)
\( \Rightarrow x = \frac{\sqrt{3}\sqrt{4-x^2}}{4} + \frac{x}{4} \)
\( \Rightarrow 4x - x = \sqrt{3}\sqrt{4-x^2} \)
\( \Rightarrow 3x = \sqrt{3}\sqrt{4-x^2} \)
Squaring both sides:
\( 9x^2 = 3(4-x^2) \)
\( \Rightarrow 9x^2 = 12 - 3x^2 \Rightarrow 12x^2 = 12 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \)
Substituting \( x = -1 \) in the original equation:
\( \cos^{-1}(-1) + \sin^{-1}(-1/2) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \neq \frac{\pi}{6} \).
Substituting \( x = 1 \):
\( \cos^{-1}(1) + \sin^{-1}(1/2) = 0 + \frac{\pi}{6} = \frac{\pi}{6} \).
\( \therefore x = 1 \).

Question. Prove that: \( \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right) = \frac{\pi}{2} \)
Answer: L.H.S. \( = \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \sin^{-1}\left( \frac{4}{5} \sqrt{1 - \left(\frac{5}{13}\right)^2} + \frac{5}{13} \sqrt{1 - \left(\frac{4}{5}\right)^2} \right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \sin^{-1}\left( \frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5} \right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \sin^{-1}\left( \frac{48 + 15}{65} \right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \sin^{-1}\left(\frac{63}{65}\right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \cos^{-1}\left(\sqrt{1 - \left(\frac{63}{65}\right)^2}\right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \cos^{-1}\left(\frac{16}{65}\right) + \sin^{-1}\left(\frac{16}{65}\right) \)
\( = \frac{\pi}{2} = \text{R.H.S.} \) (using \( \sin^{-1}z + \cos^{-1}z = \frac{\pi}{2} \))

Question. Prove that: \( 2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{4}{7} \)
Answer: L.H.S. \( = 2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} \)
\( = \tan^{-1}\left( \frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} \right) + \tan^{-1}\frac{1}{8} \)
\( = \tan^{-1}\left( \frac{2/5}{24/25} \right) + \tan^{-1}\frac{1}{8} \)
\( = \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\frac{1}{8} \)
\( = \tan^{-1}\left( \frac{\frac{5}{12} + \frac{1}{8}}{1 - \frac{5}{12}\cdot\frac{1}{8}} \right) = \tan^{-1}\left( \frac{13/24}{91/96} \right) \)
\( = \tan^{-1}\left( \frac{13}{24} \cdot \frac{96}{91} \right) = \tan^{-1}\left(\frac{4}{7}\right) = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x \text{, } 0 < x < 1 \).
Answer: \( \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x \)
\( \Rightarrow \tan^{-1}(1) + \tan^{-1}x = \frac{\pi}{4} + \tan^{-1}x \)
\( \Rightarrow \frac{\pi}{4} + \tan^{-1}x = \frac{\pi}{4} + \tan^{-1}x \)
This identity holds for all \( x \) in the domain.
Given the condition \( 0 < x < 1 \), we have \( x \in (0, 1) \).

Question. Write into the simplest form: \( \cot^{-1}\left(\sqrt{1+x^2}-x\right) \)
Answer: Let \( y = \cot^{-1}\left(\sqrt{1+x^2}-x\right) \)
Let \( x = \cot\theta \Rightarrow \theta = \cot^{-1}x \)
\( \therefore y = \cot^{-1}\left(\sqrt{1+\cot^2\theta}-\cot\theta\right) \)
\( = \cot^{-1}\left(\csc\theta - \cot\theta\right) = \cot^{-1}\left(\frac{1 - \cos\theta}{\sin\theta}\right) \)
\( = \cot^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) \)
\( = \cot^{-1}\left(\tan\frac{\theta}{2}\right) = \cot^{-1}\left(\cot\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \)
\( = \frac{\pi}{2} - \frac{\theta}{2} = \frac{\pi}{2} - \frac{1}{2}\cot^{-1}x \)

Question. Prove that: \( \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right) = \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \)
Answer: L.H.S. \( = \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right) \)
\( = \frac{9}{4}\left[ \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right) \right] \)
\( = \frac{9}{4}\cos^{-1}\left(\frac{1}{3}\right) \) (using \( \sin^{-1}z + \cos^{-1}z = \frac{\pi}{2} \))
Let \( \cos^{-1}\left(\frac{1}{3}\right) = \theta \Rightarrow \cos\theta = \frac{1}{3} \)
\( \sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \)
\( \Rightarrow \theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \)
\( \therefore \text{L.H.S.} = \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) = \text{R.H.S.} \)