# 26. Inverse Circular Functions Solutions¶

1. Let $\tan^{-1}(-1) = \theta$ then $\tan\theta = -1.$ Also, $-\frac{\pi}{2}< \theta < \frac{\pi}{2}$

The only value in this range which satisfied this equation is $-\frac{\pi}{4}.$

2. Let $\cot^{-1}(-1) = \theta,$ then $\cot\theta = -1$ and $0<\theta<\pi$

$\Rightarrow \theta = \frac{3\pi}{4}$

3. Let $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \theta$ then $\sin\theta = -\frac{\sqrt{3}}{2}$

Also, $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} \Rightarrow \theta = -\frac{\pi}{3}$

4. $\sin\left[\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2}\right)\right]$

$=\sin\left[\frac{\pi}{3} -\{-\sin^{-1}\frac{1}{2}\}\right]$

$= \sin\left[\frac{\pi}{3} + \frac{\pi}{6}\right] = 1$

5. $\sin\left[\cos^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\pi - \cos^{-1}\frac{1}{2}\right]$

$= \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$

6. $\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]$

$= \sin\left[-\frac{\pi}{3} + \pi- \frac{\pi}{6}\right]$

$= \sin\frac{\pi}{2} = 1$

7. Given expression is $\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]$

Let $\cos^{-1}\frac{\sqrt{5}}{3} = 2\theta$ then $\cos2\theta = \frac{\sqrt{5}}{3}$

$\Rightarrow \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{\sqrt{5}}{3}$

By componendo and dividendo $\frac{2\tan^2\theta}{2} = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$

$\Rightarrow \tan\theta = \pm\left(\frac{3 - \sqrt{5}}{2}\right)$

Given $0 \leq 2\theta \leq \pi \Rightarrow 0\leq \theta \leq \frac{\pi}{2}$

i.e. $\theta$ lies in first quadrant. $\Rightarrow \tan\theta = \frac{3 - \sqrt{5}}{2}$

8. Given expression is $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$

Let $\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \theta$

then $\sin\theta = \sin\frac{2\pi}{3}$ and $-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}$

$=\sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3}$

$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}$

9. Let $\sin^{-1}\frac{\sqrt{3}}{2} = \theta \Rightarrow \sin\theta = \frac{\sqrt{3}}{2}$

$\Rightarrow \theta = \frac{\pi}{3}$ $-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}$

10. Let $\tan^{-1}\frac{-1}{\sqrt{3}} =\theta$

then $\tan\theta = \frac{-1}{\sqrt{3}} =\tan\left(\frac{-\pi}{3}\right)$

$\Rightarrow \theta = -\frac{\pi}{3}$

11. Let $\cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3}$

$\cot\theta = \cot\left(\pi - \frac{\pi}{6}\right)$

$\theta = \frac{5\pi}{6}$

12. Let $\cot^{-1}\cot\frac{5\pi}{4} = \theta \Rightarrow \cot\theta = \cot\left(\pi + \frac{\pi}{4}\right)$

$\cot\theta = \cot\frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{4}$

13. Let $\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \theta$

$\Rightarrow \tan\theta = \tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right)$

$\Rightarrow \theta = -\frac{\pi}{4}$

14. $\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$

15. Let $\tan^{-1} = \theta \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \cos\theta = \frac{4}{5}$

16. $\cos\left[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right]$

$=\cos\left[\frac{\pi}{6} + \frac{\pi}{6}\right] = \cos \frac{\pi}{3} = \frac{1}{2}$

17. L.H.S. $= 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}$

$= \tan^{-1}\frac{2*\frac{1}{3}}{1 - \frac{1}{9}} + \tan^{-1}\frac{1}{7}$

$= \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7}$

$= \tan^{-1}\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}} = \tan^{-1}1 = \frac{\pi}{4} =$ R.H.S.

18. L.H.S. $= \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}$

$= \tan^{-1}\frac{\frac{1}{3} + \frac{1}{7}}{1 - \frac{1}{21}} + \tan^{-1}\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{40}}$

$= \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}1 = \frac{\pi}{4} =$ R.H.S.

19. L.H.S. $= \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}$

$= \sin^{-1}\left(\frac{4}{5}\sqrt{1 - \frac{25}{169}} + \frac{5}{13}\sqrt{1 - \frac{16}{25}}\right) + \sin^{-1}\frac{16}{65}$

$= \sin^{-1}\left(\frac{48}{65} + \frac{15}{65}\right) + \sin^{-1}\frac{16}{65} = \sin^{-1}\frac{64}{65} + \sin^{-1}\frac{16}{65}$

$= \sin^{-1}\left(\frac{63}{65}.\sqrt{1 - \frac{16^2}{65^2}} + \frac{16}{65}\sqrt{1 - \frac{63^2}{65^2}}\right)$

$\sin^{-1}\frac{63^2 + 16^2}{65^2} = \sin^{-1} = \frac{\pi}{2} =$ R.H.S.

20. We have to prove that $4\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}$

$\Rightarrow 4\tan^{-1}\frac{1}{5} = \frac{\pi}{4} + \tan^{-1}\frac{1}{70} - tan^{-1}\frac{1}{99}$

L.H.S. $= 4\tan^{-1}\frac{1}{5} = 2\tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{25}} = 2\tan^{-1}\frac{5}{12}$

$= \tan^{-1}\frac{2.\frac{5}{12}}{1 - \frac{25}{144}} = \tan^{-1}\frac{120}{119}$

$\tan^{-1}\frac{1}{70} - \tan^{-1}\frac{1}{99} = \tan^{-1}\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{70}.\frac{1}{99}}$

$= \tan^{-1}\frac{1}{239}$

R.H.S. $= \tan^{-1}1 + \tan^{-1}\frac{1}{239} = \tan^{-1}\frac{120}{119}$

Thus, L.H.S. = R.H.S.

21. We have to prove that $\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}$

$\cot^{-1}9 = \tan^{-1}\frac{1}{9}$

Let $\cosec^{-1}\frac{\sqrt{41}}{4} = \theta \Rightarrow \cosec\theta = \frac{\sqrt{41}}{4}$

Since we have to consider principal values only $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}$ and $\theta \neq 0$

As $\cosec\theta$ is +ve here, $\theta$ lies between $0$ and $\pi/2,$ hence $\tan\theta$ must be positive.

$\Rightarrow \tan\theta = \frac{4}{5}$

$\tan^{-1}\frac{1}{9} + \tan^{-1}\frac{4}{5} = \tan^{-1}\frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9}\frac{4}{5}}$

$=\tan^{-1}\frac{41}{45}.\frac{45}{41} = \tan^{-1}1 = \frac{\pi}{4} =$ R.H.S.

22. We have to prove that $4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi$

$\cot^{-1}3 = \tan^{-1}\frac{1}{3}$

$\cosec^{-1}\sqrt{5} = \tan^{-1}\frac{1}{2}$

L.H.S. $=4(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2}) = 4\left(\tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}\right)$

$= 4\tan^{-1}1 = \pi =$ R.H.S.

23. We have to prove that $\tan^{-1}x = 2\tan^{-1}[\cosec\tan^{-1}x - \tan\cot^{-1}x]$

R.H.S. $= 2\tan^{-1}[\cosec\cosec^{-1}\frac{\sqrt{1 + x^2}}{x} - \tan\tan^{-1}\frac{1}{x}]$

$= 2\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)$

Let $x = \tan\theta,$ then R.H.S. $= 2\tan^{-1}\left(\frac{sec\theta - 1}{\tan\theta}\right)$

$=2\tan^{-1}\left(\frac{1 - \cos\theta}{\sin\theta}\right) = 2\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)$

$= 2\tan^{-1}.\tan\frac{\theta}{2} = \theta = tan^{-1}x =$ L.H.S.

24. $\because 0 < b \leq a \therefore \sqrt{\frac{a - b}{a + b}}$ is real.

L.H.S. $= 2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right]$

$= \cos^{-1}\left[\frac{1 - \frac{a- b}{a + b}\tan^2\frac{x}{2}}{1 + \frac{a - b}{a + b}\tan^2\frac{x}{2}}\right]$

$= \cos^{-1}\left[\frac{a\left(1 - \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}{a\left(1 + \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}\right]$

$= \cos^{-1}\left[\frac{a\left(\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2} + b}\right)}{a + b\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}}\right]$

$= \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right] =$ R.H.S.

25. L.H.S $= \tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx}$

$= \tan^{-1}x - \tan^{-1}y + \tan^{-1}y - \tan^{-1}z + \tan^{-1}z - \tan^{-1}x = 0$

R.H.S. $= \tan^{-1}\left(\frac{x^2 - y^2}{1 + x^2y^2}\right) + \tan^{-1}\left(\frac{y^2 - z^2}{1 + y^2z^2}\right) + \tan^{-1}\left(\frac{z^2 - x^2}{1 + z^2x^2}\right)$

$= \tan^{-1}x^2 - \tan^{-1}y^2 + \tan^{-1}y^2 - \tan^{-1}z^2 + \tan^{-1}z^2 - \tan^{-1}x^2 = 0$

$\therefore$ L.H.S. = R.H.S.

26. We have to prove that $\sin\cot^{-1}\tan\cos^{-1}x = x$

L.H.S. $= \sin\cot^{-1}\tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x}$

$= \sin\cot^{-1}\frac{\sqrt{1 - x^2}}{x}$

Let $\cot^{-1}\frac{1 - x^2}{x} = \theta$ then $\cot\theta = \frac{\sqrt{1 - x^2}}{x}$

$\sin\theta = x$

Thus, L.H.S. = R.H.S.

27. Case I: When $\frac{\pi}{4}

$0<\cot x< 1$ and $0<\cot^3x<1 \therefore 0<\cot x\cot^3x<1$

$\tan^{-1}\cot x + \tan^{-1}\cot^3x =\tan^{-1}\frac{\cot x + \cot^3x}{1 - \cot x\cot^3x}$

$=\tan^{-1}\frac{\cot x}{1 - \cot^2x} = \tan^{-1}\frac{\tan x}{\tan^2 - 1}$

$= -\tan^{-1}\left(\frac{1}{2}\tan 2x\right)$

$\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = 0$

Case II: When $0

$\cot x> 1$ and $\cot^3x > 1$

$\Rightarrow \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi - \tan^{-1}\left(\frac{1}{2}\tan2x\right)$

$\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi$

28. $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}$

$= \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}1 = \frac{\pi}{4}$

$\tan^{-1}\frac{3}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5}.\frac{1}{4}}$

$=\tan^{-1}\frac{17/20}{17/20} = \tan^{-1}1 = \frac{\pi}{4}$

29. We have to prove that $\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}$

L.H.S. $= \tan^{-1}\frac{\frac{2a - b}{\sqrt{3}b} + \frac{2b - a}{\sqrt{3}a}}{1 - \frac{(2a - b)(2b - a)}{3ab}}$

$= \tan^{-1}\frac{\frac{2\sqrt{3}a^2 - \sqrt{3}ab + 2\sqrt{3}b^2 -\sqrt{3}ab}{3ab}}{\frac{3ab - 4ab + 2a^2 + 2b^2 - ab}{3ab}}$

$= \tan^{-1}\frac{2\sqrt{a^2} + 2\sqrt{3}b^2 - 2\sqrt{3}ab}{2a^2 + 2b^2 - 2ab} = \tan^{-1}\sqrt{3} = \frac{\pi}{3}$

30. We have to prove that $\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}$

L.H.S. $= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{\frac{1}{3} + \frac{1}{12}}{1 - \frac{1}{3}.\frac{1}{12}}$

$= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{5/12}{35/36} = \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{3}{7}$

$= \tan^{-1}\frac{\frac{2}{5} + \frac{3}{7}}{1 - \frac{2}{5}.\frac{3}{7}}$

$= \tan^{-1}\frac{29/35}{29/35} = \tan^{-1}1 = \frac{\pi}{4}$

31. We have to prove that $2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}$

L.H.S. $= \tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{5^2}} + \tan^{-1}\frac{1}{4}$

$= \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{4}$

$= \tan^{-1}\frac{\frac{5}{12} + \frac{1}{4}}{1 - \frac{5}{12}.\frac{1}{4}}$

$= \tan^{-1}\frac{2/3}{43/48} = \tan^{-1}\frac{32}{43}$

32. We have to prove that $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)$

$\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}1 + \tan^{-1}\frac{2 + 3}{1 - 2.3} = \tan^{-1}1 + \tan^{-1}(-1)$

$= \tan^{-1}\frac{1 - 1}{1 + 1} =\tan^{-1}0 = n\pi$

$2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)$

$= 2\left(\tan^{-1}1 + \tan^{-1}\frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2}.\frac{1}{3}}\right)$

$= 2\left(\frac{\pi}{4} + \tan^{-1}1\right) = 2.\frac{\pi}{2} = \pi$

Thus, the above expression will have principal value as $pi.$

33. We have to prove that $\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}$

L.H.S. $= \tan^{-1}x + \cot^{-1}y = \tan^{-1}x + \tan^{-1}\frac{1}{y}$

$= \tan^{-1}\frac{x + \frac{1}{y}}{1 - x.\frac{1}{y}} = \tan^{-1}\frac{xy + 1}{y - x}$

34. We have to prove that $\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x$

L.H.S. $= \tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1}$

$= \tan^{-1}\frac{\frac{1}{x + y} + \frac{y}{x^2 + xy + 1}}{1 - \frac{1}{x + y}.\frac{y}{x^2 + xy + 1}}$

$= \tan^{-1}\frac{x^2 + 2xy + y^2 + 1}{x^3 + 2x^2y + xy^2 + x} = \tan^{-1}\frac{1}{x} = \cot^{-1}x$

35. We have to prove that $2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4$

We know that $\cot^{-1}x + \cot^{-1}y = \frac{xy - 1}{x + y}$

$\therefore 2\left(\cot^{-1}5 + \cot^{-1}8\right) = 2\cot^{-1}\frac{39}{13} = 2\cos^{-1}3 = \cot^{-1}\frac{4}{3}$

$\therefore 2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \cot^{-1}\frac{4}{3} + \cot^{-1}7$

$=\cot^{-1}\frac{\frac{28}{3} - 1}{\frac{25}{3}} = \cot^{-1}1 = \pi/4$

36. We have to prove that $\tan^{-1}\frac{a - b}{1 + ab} + \tan^{-1}\frac{b - c}{1 + bc} + \tan^{-1}\frac{c - a}{1 + ca} = 0$

L.H.S. $= \tan^{-1}a - \tan^{-1}b + \tan^{-1}b - \tan^{-1}c + \tan^{-1}c - \tan^{-1}a = 0$

37. We have to prove that $\tan^{-1}\frac{a^3 - b^3}{1 + a^3b^3} + \tan^{-1}\frac{b^3 - c^3}{1 + b^3c^3} + \tan^{-1}\frac{c^3 - a^3}{1 + c^3a^3} = 0$

L.H.S. $= \tan^{-1}a^3 - \tan^{-1}b^3 + \tan^{-1}b^3 - \tan^{-1}c^3 + \tan^{-1}c^3 - \tan^{-1}a^3 = 0$

38. We have to prove that $\cot^{-1}\frac{xy + 1}{y - x} + \cot^{-1}\frac{yz + 1}{z - y} + \cot^{-1}z = \tan^{-1}\frac{1}{x}$

L.H.S. $= \cot^{-1}x - \cot^{-1}y + \cot^{-1}y - \cot^{-1}z + \cot^{-1}z= \cot^{-1}x = \tan^{-1}\frac{1}{x}$

39. We have to prove that $\cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)$

L.H.S. $= \cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right)$

$= \tan^{-1}\frac{\sqrt{1 + \cos^2\theta\cos^2\phi + 2\cos\theta\cos\phi - \cos^2\theta\cos^2\phi - 2\cos\theta\cos\phi}}{\cos\theta + \cos\phi}$

$= \tan^{-1}\frac{\sqrt{(1 - \cos^2\theta)(1 - \cos^2\phi)}}{\cos\theta + \cos\phi} = \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}$

R.H.S. $= 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)$

$= \tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 - \tan^2\frac{\theta}{2}\tan^2\frac{\phi}{2}}$

$=\tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}.\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2}}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \sin^2\frac{\theta}{2}\sin^2\frac{\phi}{2}}$

$= \tan^{-1}\frac{1}{2}.\frac{\sin\theta\sin\phi}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \left(1 - \cos^2\frac{\theta}{2}\right)\left(1-\cos^2\frac{\phi}{2}\right)}$

$= \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}$

40. We have to prove that $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}$

L.H.S. $= \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17}$

$=\sin^{-1}\left(\frac{3}{5}\sqrt{1 - \frac{8^2}{17^2}} + \frac{8}{17}\sqrt{1 - \frac{3^2}{5^2}}\right)$

$= \sin^{-1}\left(\frac{3}{5}.\frac{15}{17} + \frac{8}{17}.\frac{4}{5}\right)$

$\sin^{-1}\left(\frac{45 + 32}{85}\right) = \sin^{-1}\frac{77}{85} =$ R.H.S.

41. We have to prove that $\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}$

We know that $\cos^{-1}x + \cos^{-1}y = xy - \sqrt{(1 - x^2)(1 - y^2)}$

L.H.S. $= \cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65}$

$= \cos^{-1}\left(\frac{3}{5}.\frac{12}{13} - \sqrt{\left(1 - \frac{3^2}{5^2}\right)\left(1 - \frac{12^2}{13^2}\right)}\right) + \cos^{-1}\frac{63}{65}$

$= \cos^{-1}\left(\frac{36}{65} - \frac{4}{5}.\frac{5}{13}\right) + \cos^{-1}\frac{63}{65}$

$= \cos^{-1}\left(\frac{36}{65} - \frac{20}{65}\right) + \cos^{-1}\frac{63}{65}$

$= \cos^{-1}\frac{16}{65} + \cos^{-1}\frac{63}{65}$

$= \cos^{-1}\left(\frac{16}{65}.\frac{63}{64} - \sqrt{\left(1 - \frac{16^2}{65^2}\right)\left(1 - \frac{63^2}{65^2}\right)}\right)$

$= \cos^{-1}0 = \frac{\pi}{2} =$ R.H.S.

42. We have to prove that $\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right)$

L.H.S. $= \sin^{-1}x + \sin^{-1}y = \cos^{-1}\sqrt{1 - x^2} + \cos^{-1}\sqrt{1 - y^2}$

$= \cos^{1-}(\sqrt{1 - x^2}\sqrt{1 - y^2} - \sqrt{[1 - (1 - x^2)][1 - (1 - y^2)]})$

$= \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right) =$ R.H.S.

43. We have to prove that $4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi$

or $\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}} =\pi/4$

L.H.S. $={\mathrm{sin}}^{-1}\frac{1}{\sqrt{10}}+{}^{}$