Inverse Trigonometric Functions

Class 12 NCERT

NCERT

1   Find the principal value of $\sin^{-1} (-\dfrac{1}{2})$

Solution :

Let $\sin^{-1}(-\dfrac{1}{2})=y$,Then $sin y =-\dfrac{1}{2}=-\sin (\dfrac{\pi}{6}) =\sin(-\dfrac{\pi}{6})$$\\$ We know that the range of the principal value branch of $\sin^{-1} $ is $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ and $\sin(-\dfrac{\pi}{6}) =\dfrac{1}{2},$ $\\$ Therefore, the principal value of $\sin^{-1} (-\dfrac{1}{2}) $ is $-\dfrac{\pi}{6}$

2   Find the principal value of $\cos^{-1} (\dfrac{\sqrt{3}}{2})$

Solution :

Let $\cos ^{-1} (\dfrac{\sqrt{3}}{2})=y$$\\$ Then $\cos y =\dfrac{\sqrt{3}}{2}=\cos (\dfrac{\pi}{6})$$\\$ We know that the range of the principal value branch of $\cos ^{-1}$ is [$0,\pi $]$\\$ and $\cos (\dfrac{\pi}{6}) =\dfrac{\sqrt{3}}{2}.$$\\$ Therefore, the principal value of $\cos ^{-1} (\dfrac{\sqrt{3}}{2}) $ is $\dfrac{\pi}{6}$

3   Find the principal value of $\csc ^{-1} (2)$

Solution :

Let $\csc^{-1}(2)=y$$\\$ Then ,$\csc y=2=\csc(\dfrac{\pi}{6})$$\\$ We know that the range of the principal value branch of $\csc ^{-1} $ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}] -\{0\}$$\\$ Therefore ,the principal value of $\csc ^{-1}(2) $ is $\dfrac{\pi}{6}$

4   Find the principal value of $\tan ^{-1} (-\sqrt{3})$

Solution :

Let $\tan ^{-1}(-\sqrt{3})=y$$\\$ Then ,$\tan y=-\sqrt{3}=-\tan \dfrac{\pi}{3} =\tan({-\dfrac{\pi}{3}})$$\\$ We know that the range of the principal value branch of $\tan^{-1}$ is $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ and $\tan (-\dfrac{\pi}{3}) $ is $-\sqrt{3}.$$\\$ Therefore, known that the principal value of $\tan^{-1}(-\sqrt{3}) $ is $-\dfrac{\pi}{3}.$

5   Find the principal value of $\cos^{-1}(-\dfrac{1}{2})$

Solution :

Let $\cos^-1(-\dfrac{1}{2})=y$$\\$ Then , $\cos y=-\dfrac{1}{2} =-\cos(\dfrac{\pi}{3})=\cos (\pi-\dfrac{\pi}{3})=\cos (\dfrac{2\pi}{3}).$$\\$ We know that the range of the principal value branch of $\cos^{-1}$ is $[0,\pi]$$\\$ and $\cos (\dfrac{2\pi}{3})=-\dfrac{1}{2}$$\\$ Therefore , the principal value of $\cos^{-1}(-\dfrac{1}{2})$ is $(\dfrac{2\pi}{3})$

6   Find the principal value of $\tan^{-1}(-1)$

Solution :

Let $ \tan^{-1}(-1)=y$$\\$ Then ,$\tan y=-1 =-\tan (\dfrac{\pi}{4}) =\tan (-\dfrac{\pi }{4}).$$\\$ We know that the range of the principle value branch of $\tan^{-1} $ is $ (-\dfrac{\pi}{2},\dfrac{\pi}{2})$ and $\tan{-\dfrac{\pi}{4}}=-1.$$\\$ Therefore , the principle value of $ \tan^{-1 }(-1) $ is $ -\dfrac{\pi}{4 } $

7   Find the principal value of $ \sec^{-1} (\dfrac{2}{\sqrt{3}})$

Solution :

Let ,$sec^{-1} (\dfrac{2}{\sqrt{3}})=y .$ Then $ \sec y =\dfrac{2}{\sqrt{3}} =\sec(\dfrac{\pi}{6}) .$ $\\$ We know that the range of the principal value branch of $\sec^{-1}$ is $[0.\pi] -{\dfrac{\pi}{2}}$ and $\sec(\dfrac{\pi}{6})=\dfrac{2}{\sqrt{3}}.$$\\$ Therefore , the principle value of $ \sec^{-1} (\dfrac{2}{\sqrt{3}}) $ is $\dfrac{\pi}{6}.$

8   Find the principal value of $\cot ^{-1} (\sqrt{3})$

Solution :

Let $ \cot ^{-1}(\sqrt{3})=y.$ $\\$ Then $\cot y=\sqrt{3} =\cot(\dfrac{\pi}{6}).$$\\$ We know that the range of the principal value branch of $\cot^{-1} $ is $(0,\pi)$ and $\cot \dfrac {\pi}{6} =\sqrt{3}.$$\\$ Therefore , the principle value of $ \cot^{-1} (\sqrt{3}) $ is $ \dfrac{\pi}{6}$

9   Find the principal value of $\cos ^{-1} (-\dfrac{1}{\sqrt{2}})$

Solution :

Let $ \cos ^{-1}(-\dfrac{1}{\sqrt{2}}) =y$$\\$ Then $ \cos y =-\dfrac{1}{\sqrt{2}} =-\cos (\dfrac{\pi}{4}) \\ =\cos (\pi -\dfrac{\pi}{4}) = cos (\dfrac{3\pi}{4}).$$\\$ We know that the range of the principal value branch of $ \cos ^{-1} $ is $[0,\pi]$ and $ \cos (\dfrac{3\pi}{4})=-\dfrac{1}{\sqrt{2}}.$$\\$ Therefore , the principle value of $ \cos^{-1}(-\dfrac{1}{\sqrt{2}})$ is $\dfrac{3\pi}{4}.$

10   Find the principal value of $\csc ^{-1}(-\sqrt{2})$

Solution :

Let $\csc^{-1} (-\sqrt{2})=y $. Then , $ \csc y=-\sqrt{2} =-\csc(\dfrac{\pi}{4}) =\csc(-\dfrac{\pi}{4}).$$\\$ We know that the range of the principle value branch of $ \csc^{-1}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]-{0}$ and $ \csc(-\dfrac{-\pi}{4}) =-\sqrt{2}. $$\\$ Therefore , the principle value of $ \csc^{-1}(-\sqrt{2}) $ is $-\dfrac{\pi}{4}.$

11   Find the value of $\tan ^{-1}(1)+ \cos ^{-1}(-\dfrac{1}{2}) +\sin^{-1} (-\dfrac{1}{2})$

Solution :

Let $\tan^{-1}(1)=x$ $\\$ Then,$\tan x =1=\tan(\dfrac{\pi}{4}).$ $\\$ $\therefore \tan^{-1} (1) =\dfrac{\pi}{4}$ $\\$ Let $ \cos^{-1}(-\dfrac{1}{2})=y$ $\\$ Then , $\cos y =-\dfrac{1}{2} =-\cos (\dfrac{\pi}{3})\\ =\cos (\pi -\dfrac{\pi}{3}) =\cos (\dfrac{2\pi}{3}).$ $\\$ $\therefore \cos ^{-1} (-\dfrac{1}{2}) =\dfrac{2\pi}{3}$$\\$ Let $ \sin^{-1}(-\dfrac{1}{2}) =z.$$\\$ Then , $\sin z =-\dfrac{1}{2} =-\sin \dfrac{\pi}{6}\\ \sin (-\dfrac{\pi}{6})\\ \therefore \sin^{-1}(-\dfrac{1}{2}) =-\dfrac{\pi}{6}\\ \therefore \tan^{-1}(1) + \cos^{-1} (-\dfrac{1}{2}) + \sin^{-1}(-\dfrac{1}{2}) \\ =\dfrac{\pi }{4}+\dfrac{2\pi}{3} -\dfrac{\pi}{6}\\ =\dfrac{3\pi +8\pi- 2\pi}{12}=\dfrac{9\pi}{12}=\dfrac{3\pi}{4}$

12   Find the value of $ \cos^{-1}(\dfrac{1}{2})+ 2\sin^{-1}(\dfrac{1}{2})$

Solution :

Let $ \cos^{-1} (\dfrac{1}{2})=x.$$\\$ Then , $\cos x=\dfrac{1}{2} =\cos(\dfrac{\pi}{3}).$$\\$ $\therefore \cos ^{-1}(\dfrac{1}{2}) =\dfrac{\pi}{3}$$\\$ Let $ \sin^{-1}(\dfrac{1}{2}) =y $ $\\$ Then , $\sin y =\dfrac{1}{2} =\dfrac{\pi}{6} .$$\\$ $\therefore \sin^{-1} (\dfrac{1}{2}) =\dfrac{\pi}{6}\\ \therefore \cos^{-1} (\dfrac{1}{2}) + 2 \sin ^{-1} (\dfrac{1}{2})\\ =\dfrac{\pi} {3}+2*\dfrac{\pi}{6} =\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$

13   If $ \sin^{-1}x=y , $ then $\\$ $(A) 0 \leq y \leq \pi \\ (B)-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}\\ (C) 0 < y < \pi \\ (D)-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$

Solution :

It is given that $ \sin^{-1} x=y $$\\$ We know that the range of the principal value branch of $ \sin^{-1} $ is $ [-\dfrac{\pi}{2},\dfrac{\pi}{2}].$$\\$ Therefore , $ -\dfrac{ \pi}{2} \leq y \leq \dfrac{\pi}{2}$$\\$ Answer choise $(B)$ is correct

It is given that $ \sin^{-1} x=y $$\\$ We know that the range of the principal value branch of $ \sin^{-1} $ is $ [-\dfrac{\pi}{2},\dfrac{\pi}{2}].$$\\$ Therefore , $ -\dfrac{ \pi}{2} \leq y \leq \dfrac{\pi}{2}$$\\$ Answer choise $(B)$ is correct

14   $ \tan^{-1} \sqrt{3}-\sec^{-1}(-2) $ is equal to

Solution :

Let $\tan^{-1} \sqrt{3} =x$$\\$ Then , $\tan x =\sqrt{3}=\tan\dfrac{\pi}{3}$$\\$ We know that the range of the principal value branch of $\tan^{-1} $ is $(\dfrac{-\pi}{2},\dfrac{\pi}{2}).$$\\$ $\therefore \tan^{-1} \sqrt{3} =\dfrac{\pi}{3}$$\\$ Let $\sec^{-1} (-2) =y $$\\$ Then , $ \sec y=-2 =-\sec(\dfrac{\pi}{3}) =\sec(\pi-\dfrac{\pi}{3}) =\sec\dfrac{2\pi}{3}.$$\\$ We know that the range of the principal value branch of$\sec^{-1} $ is $ [0,\pi]-{\dfrac{\pi}{2}}.\\ \therefore \sec^{-1}(-2)=\dfrac{2\pi}3 $$\\$ Thus ,$ \tan^{-1} (\sqrt{3}) -\sec^{-1}(-2)\\ =\dfrac{\pi}{3}-\dfrac{2\pi}{3} =-\dfrac{\pi}{3}$

15   Prove $ 3\sin^{-1} x=\sin^{-1}(3x-4x^3),x \in [-\dfrac{1}{2},\dfrac{1}{2}]$

Solution :

To prove $ 3\sin ^{-1} x=\sin^{-1}(3x -4x^3) ,$ where $ x\in [-\dfrac{1}{2},\dfrac{1}{2}] $$\\$ Let $ x=\sin \theta $ Then $ \sin^{-1} x=\theta $$\\$ We have ,$\\$ R.H.S. $\\$ $\sin^{-1} (3x -4x^3) =\sin^{-1} (3\sin \theta -4 \sin^3\theta) \\ =\sin^{-1} (\sin3\theta )\\ 3 \theta \\ = 3 \sin^{-1} = $ L.H.S

16   Prove $3\cos^{-1} x =\cos^{-1}(4x^3-3x) , x\in [\dfrac{1}{2},1]$

Solution :

17   Prove $3\cos^{-1} x =\cos^{-1}(4x^3-3x) , x\in [\dfrac{1}{2},1]$

Solution :

To prove $3\cos^{-1} x=\cos^{-1}(4x^3-3x),x\in [\dfrac{1}{2},1]$$\\$ Let,$x=\cos \theta . $ Then $\cos ^{-1} x=\theta $$\\$ We have R.H.S. $ \cos^{-1}(4x^3- 3x)\\ =\cos^{-1}(4\cos^3\theta -3 \cos \theta )\\ =\cos^{-1}(\cos 3 \theta )\\ 3 \theta \\ = 3\cos^{-1} x =$ L.H.S.

18   Prove $\tan^{-1} (\dfrac{2}{11}) +\tan^{-1} (\dfrac{7}{24}) =\tan^{-1}(\dfrac{1}{2})$

Solution :

To prove: $ \tan^{-1} \dfrac{2}{11} + \tan^{-1}\dfrac{7}{24} =\tan^{-1}\dfrac{1}{2}$$\\$ L.H.S. $\\$ $ \tan^{-1}\dfrac{2}{11} +\tan^{-1}\dfrac{7}{24}\\ = \tan^{-1} \dfrac{\dfrac{2}{11}+\dfrac{7}{11}}{1-(\dfrac{2}{11}.\dfrac{7}{24})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}]\\ =\tan^{-1}\dfrac{11*24}{\dfrac{11*24-14}{11*24}} \\ =\tan^{-1}\dfrac{48+77}{264-14}\\ =\tan^{-1}(\dfrac{125}{250}) \\ =\tan^{-1}(\dfrac{1}{2}) = $ R.H.S.

19   Prove $ 2\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{7}=\tan^{-1}\dfrac{31}{17}$

Solution :

$\tan^{-1}\dfrac{1}{\dfrac{3}{4}}+\tan^{-1}\dfrac{1}{7}\\ =\tan^{-1}\dfrac{4}{3}+\tan^{-1}\dfrac{1}{7}\\ =\tan^{-1}\dfrac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}.\frac{1}{7}}[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}]\\ =\tan^{-1}(\dfrac{\frac{28+3}{21}}{\frac{21-4}{21}})\\ =\tan^{-1}\dfrac{31}{17}$= R.H.S.

To Prove $ 2\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{7}=\tan^{-1}\dfrac{31}{17}$$\\$ L.H.S. =$ 2 \tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{7}\\ =\tan^{-1}\dfrac{2.\dfrac{1}{2}}{1-(\dfrac{1}{2})^2} + \tan^{-1}\dfrac{1}{7}[2\tan^{-1}x =\tan^{-1}\dfrac{2x}{1-x^2}]$

20   Write the function in the simplest form:$\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x},x\neq 0$

Solution :

$\tan^{-1}(\dfrac{2\sin^2(\dfrac{\theta}{2})}{2 \sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}) \\ =\tan^{-1}(\tan\dfrac{\theta}{2})=\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x$

$\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$$\\$ Put $ x=\tan\theta \implies \theta =\tan^{-1}x\\ \therefore \tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}\\ =\tan^{-1}\dfrac{\sqrt{1+\tan^2\theta }-1}{\tan \theta } \\ =\tan^{-1}(\dfrac{\sec \theta -1}{\tan \theta }) \\ =\tan^{-1}(\dfrac{1-\cos\theta}{\sin \theta}) $$\\$ $\tan^{-1}(\dfrac{2\sin^2(\dfrac{\theta}{2})}{2 \sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}) \\ =\tan^{-1}(\tan\dfrac{\theta}{2})=\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x$

21   Write the function in the simplest form:$\tan^{-1}(\dfrac{1}{\sqrt{x^2-1}}),|x|>1$

Solution :

$\tan^{-1}\dfrac{1}{\sqrt{x^2-1}},|x|>1$$\\$ Put $ x=\csc\theta \implies \theta =\csc^{-1}x\\ \therefore \tan^{-1}\dfrac{1}{\sqrt{x^2-1}} \\ =\tan^{-1}\dfrac{1}{\sqrt{\csc^2\theta -1}}\\ =\tan^-1(\dfrac{1}{\cot\theta})\\ =\tan^{-1}(\tan\theta)\\ =\theta \\ =\csc^{-1}x\\ =\dfrac{\pi}{2}-\sec^{-1} x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [As, \csc^{-1}x+\sec^{-1}x=\dfrac{\pi}{2}]$

22   Write the function in the simplest form:$\tan^{-1}(\sqrt{\dfrac{1-\cos x}{1+\cos x}}), x<\pi $

Solution :

$\tan^{-1}(\sqrt{\dfrac{1-\cos x}{1+\cos x}}),x< \pi\\ \tan^{-1}(\sqrt{\dfrac{1-\cos x}{1+\cos x}}) \\ =\tan^{-1}(\sqrt{\dfrac{2 \sin^2\frac{x}{2}}{2 \cos^2\frac{x}{2}}}) \\ =\tan^{-1}(\dfrac{\sin\frac{x}{2}}{\cos \frac{x}{2}})\\ =\tan^{-1}(\tan \frac{x}{2})\\ \dfrac{x}{2}$

23   Write the function in the simplest form:$\tan^{-1}(\dfrac{\cos x-\sin x}{\cos x+ \sin x}),0 < x < \pi $

Solution :

$\tan^{-1}(\dfrac{\cos x-\sin x}{\cos x + \sin x})\\ =\tan^{-1}(\dfrac{1-(\dfrac{\sin x}{\cos x})}{1+(\dfrac{\sin x}{\cos x})})\\ =\tan^{-1} (\dfrac{1-\tan x}{1+\tan x})\\ =\tan^{-1}(1)-\tan^{-1}(\tan x) \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \frac{-y}{xy} =\tan^{-1} x -\tan^{-1} y]\\ =\dfrac{\pi}{4}-x$

24   Write the function in the simplest form:$\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}},|x| < a$

Solution :

$\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}}$$\\$ Let ,$ x=a \sin \theta \implies \dfrac{x}{a}=\sin \theta \implies \sin^{-1}(\dfrac{x}{a})\\ \therefore \tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}}\\ =\tan^{-1}(\dfrac{a \sin \theta}{\sqrt{a^2 -a62 \sin^2 \theta }})\\ =\tan^{-1}(\dfrac{a \sin \theta}{a\sqrt{1-\sin^2 \theta }})\\ =\tan^{-1}(\dfrac{a \sin \theta }{a \cos \theta})\\ =\tan^{-1}(\tan \theta) =\theta = \sin^{-1}\dfrac{x}{a}$

25   Write the function in the simplest form:$\tan^{-1}(\dfrac{3a^2 x- x^3}{a^3 - 3ax^2}),a > 0 ; \dfrac{-a}{\sqrt{3}} \leq x \leq \dfrac{a}{\sqrt{3}}$

Solution :

Consider, $ \tan^{-1}(\dfrac{3 a^2 x-x^3}{a^3 - 3 ax^2})$$\\$ Let , $ x= a \tan \theta \implies \dfrac{x}{a} =\tan \theta \\ \implies \theta =\tan^{-1}(\dfrac{x}{a})\\ \tan^{-1} (\dfrac{3a^2 x-x^3}{a^3 - 3 a x^2})\\ =\tan^{-1}(\dfrac{3 a^2.a \tan \theta - a^3 \tan ^3 \theta}{a^3 - 3 a. a^2 \tan^2 \theta})\\ =\tan^{-1}(\dfrac{3a^3 \tan \theta - a^3 \tan^3 \theta}{a^3- 3 a^3 \tan^2 \theta})\\ =\tan ^{-1}(\tan \theta)\\ =3 \theta \\ 3 \tan ^{-1}\dfrac{x}{a}$

26   Find the value of $ \tan^{-1}[2 \cos(2 \sin^{-1}\frac{1}{2})]$

Solution :

Let $ \sin^{-1} \dfrac{1}{2}=x$$\\$ Then $\sin x =\dfrac{1}{2}= \sin(\dfrac{\pi}{6})\\ \therefore \sin^{-1}\dfrac{1}{2}=\dfrac{\pi}{6}\\ \therefore \tan^{-1}[2cos(2\sin^{-1}\dfrac{1}{2})]\\ =\tan^{-1}[2 \cos (2*\dfrac{\pi}{6})]\\ =\tan^{-1} [2 \cos \dfrac{\pi}{3}]\\ =\tan^{-1}[2*\dfrac{1}{2}]\\ \tan^{-1} = \dfrac{\pi}{4}$

27   Find the value of $\cot (\tan^{-1} a+ \cot^{-1} a)$

Solution :

$\cot(\tan^{-1} a+\cot^{-1} a)\\ = \cot(\dfrac{\pi}{2}) \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \vdash \cot^{-1}x =\dfrac{\pi}{2}]\\ = 0$

28   Find the value of $\tan \dfrac{1}{2} [\sin^{-1}\dfrac{2x}{1+x^2}+ \cos^{-1}\dfrac{1-y^2}{1+ y^2}] ,|x| < 1, y>0$ and $ xy < 1 $

Solution :

Let $ x=\tan \theta $$\\$ Then , $ \theta = \tan^{-1} x\\ \therefore \sin^{-1}\dfrac{2 x}{1+ x^2} =\sin^{-1} (\dfrac{2 \tan \theta}{1+ \tan^2 \theta})\\ =\sin^{-1}(\sin 2 \theta)\\ =2 \theta \\ =2 \tan^{-1}x$$\\$ Let $ y=\tan \theta .$ Then $ \theta = \tan^{-1}y . $$ \therefore \cos ^{-1}(\dfrac{1-y^2}{1+y^2}) \\ =\cos ^{-1}(\dfrac{1-\tan^2 \theta}{1+\tan^2 \theta})\\ =\cos^{-1}(\cos 2 \theta)\\ =2 \theta = 2 \tan^{-1}y\\ \therefore \tan \dfrac{1}{2}[\sin^{-1}(\dfrac{2 x}{1+x^2})+\cos^{-1}(\dfrac{1-y^2}{1+y^2})]\\ =\tan\dfrac{1}{2}[2 \tan^{-1}x+ 2 \tan^{-1}y] \ \ \ \ \ \ \ \ [As,\tan^{-1}x+\tan^{-1} y = \tan^{-1}\dfrac{x+y}{1-xy}]\\ =\tan[\tan^{-1}(\dfrac{x+y}{1-xy})]\\ =\dfrac{x+y}{1-xy}$

29   If $\sin(\sin^{-1}\dfrac{1}{5}+\cos^{-1}x)=1 $ , then find the value of x.

Solution :

$\sin (\sin ^{-1}\dfrac{1}{5}+ \cos^{-1} x)=1\\ \implies \sin(\sin^{-1}\dfrac{1}{5}) \cos (\cos^{-1}x)+\cos (\sin^{-1}\dfrac{1}{5}) \sin(\cos^{-1}x)=1\\ [\therefore B) =\sin A .\cos B + \cos A . \sin B]\\ \implies \dfrac{1}{5}.x+\cos(\sin^{-1}\dfrac{1}{5})\sin(\cos^{-1}x)=1\\ \implies \dfrac{x}{5}+ \cos (\sin^{-1}\dfrac{1}{5}) \sin (\cos^{-1}x) =1 ...................(1)$$\\$ Now,let $ \sin^{-1}\dfrac{1}{5} =y $$\\$ Then,$\sin^{-1}\dfrac{1}{5}=y\\ \sin y=\dfrac{1}{5}\\ \implies \cos y =\sqrt{1-(\dfrac{1}{5})^2} =\dfrac{2\sqrt{6}}{5}\\ \implies y= cos ^{-1}(\dfrac{2\sqrt{6}}{5})\\ \therefore \sin^{-1}\dfrac{1}{5} =\cos ^{-1}(\dfrac{2\sqrt{6}}{5}).................................(2)$$\\$ Let $\cos^{-1} x =z.$ Then , $\cos z =x\\ \implies \sin z = \sqrt{1-x^2}\\ \implies z=\sin^{-1}(\sqrt{1-x^2}).....................(3)$$\\$ From (1),(2) and (3) we have: $\\$ $ \dfrac{x}{5}+\cos(\cos^{-1}\dfrac{2\sqrt{6}}{5}).\sin(\sin^{-1}\sqrt{1-x^2}) =1 \\ \implies \dfrac{x}{5}+ \dfrac{2\sqrt{6}}{5}.\sqrt{1-x62}=1\\ \implies x+2\sqrt{6}\sqrt{1-x^2} =5\\ \implies 2\sqrt{6}\sqrt{1-x^2} =5-x$$\\$ On squaring both sides, we get:$\\$ $(2\sqrt{6})^2(1-x^2) =25+x^2-10 x\\ \implies (4*6)(1-x^2) =25+x^2-10 x\\ \implies 24-24x^2=25+x^2-10x\\ \implies 25x^2-10x+1=0\\ \implies (5x-1)^2=0\\ \implies(5x -1)=0\\ \implies x=\dfrac{1}{5}$ Hence , the value of $x $ is $ \dfrac{1}{5}. $

30   If $ \tan^{-1}\dfrac{x-1}{x-2}+\tan^{-1}\dfrac{x+1}{x+2}=\dfrac{\pi}{4}$, then find the value of x.

Solution :

$\tan^{-1}\dfrac{x-1}{x-2}+\tan^{-1}\dfrac{x+1}{x+2}=\dfrac{\pi}{4}\\ \implies \tan^{-1}\left[\dfrac{\dfrac{x-1}{x-2}+\dfrac{x+1}{x+2}}{1-(\dfrac{x-1}{x+2})(\dfrac{x+1}{x+2})}\right] =\dfrac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [As,\tan^{-1}x+\tan^{-1} y=\tan^{-1}\dfrac{x+y}{1-xy}]\\ \implies \tan^{-1}\left[\dfrac{(x-1)(x+2)+(x+1)(x-2)}{(x+2)(x-2)-(x-1)(x+1)}\right] =\dfrac{\pi}{4}\\ \implies \tan^{-1}\left[\dfrac{x^2+ x-2+x^2-x-2}{x62-4-x^2+1}\right]=\dfrac{\pi}{4}\\ \implies \tan^{-1} [\dfrac{2x^2-4}{-3}] =\dfrac{\pi}{4}\\ \implies \tan[\tan^{-1}\dfrac{4-2x^2}{3}] =\tan \dfrac{\pi}{4}\\ \implies \dfrac{4-2x^2}{3}=1\\ \implies 4-2X^2 =3\\ \implies 2x^2=4-3=1\\ \implies x=\pm \dfrac{1}{\sqrt{2}}$$\\$ Hence , the value of x is $ \pm \dfrac{1}{\sqrt{2}}.$

31   Find the values of $ \sin^{-1}(\sin\dfrac{2\pi}{3})$

Solution :

Consider , $ \sin^{-1}(\sin \dfrac{2 \pi}{3})$$\\$ We know that $ \sin^{-1}(\sin x)=x$$\\$ If $ x \in [-\dfrac{\pi}{2},\dfrac{\pi}{2}],$ which is the principal value branch of $ \sin^{-1}x..$$\\$ Here , $\dfrac{2\pi}{3} \notin [-\dfrac{\pi}{2},\dfrac{\pi}{2}]$$\\$ Now , $ \sin^{-1}(\sin\dfrac{2 \pi}{3})$ can be written as: $\\$ $ \sin^{-1}(\sin\dfrac{2\pi}{3})\\ = \sin^{-1}[\sin(\pi-\dfrac{2 \pi}{3})]\\ =\sin^{-1}(\sin\dfrac{\pi}{3}),$ where $ \dfrac{\pi}{3}\in[\dfrac{-\pi}{2},\dfrac{\pi}{2}]\\ \therefore \sin^{-1}(\sin \dfrac{2\pi}{3}) =\sin^{-1}[\sin \dfrac{\pi}{3}] =\dfrac{\pi}{3}$

32   Find the values of $ \tan^{-1}(\tan \dfrac{3\pi}{4})$

Solution :

Consider , $ \tan^{-1}(\tan\dfrac{3 \pi}{4})$$\\$ We know that $ \tan^{-1}(\tan x)=x $$\\$ If $ x=\in (-\dfrac{\pi}{2},\dfrac{\pi}{2}),$ which is the principal value branch of $ \tan^{-1} x$$\\$ Here , $ \dfrac{3 \pi}{4}\notin (-\dfrac{\pi}{2},\dfrac{\pi}{2}).$ Now, $ \tan^{-1}(\tan \dfrac{3 \pi}{4})$ can be written as :$\\$ $ \tan^{-1} (\tan \dfrac{3 \pi}{4})=\tan^{-1}[-\tan(\dfrac{-3 \pi}{4})]\\ = \tan^{-1}[-\tan(\pi-\dfrac{\pi}{4})]\\ \tan^{-1}(-\tan\dfrac{\pi}{4}) =\tan^{-1}[\tan(\dfrac{-\pi}{4})] $ where $-\dfrac{\pi}{4}\in(-\dfrac{\pi}{2},\dfrac{\pi}{2})\\ \therefore \tan^{-1}(\tan \dfrac{3\pi}{4}) =\tan^{-1}[\tan(\dfrac{-\pi}{4})]=\dfrac{-\pi}{4}$

33   Find the values of $ \tan(\sin^{-1}\dfrac{3}{5}+ \cot^{-1}\dfrac{3}{2})$

Solution :

Let $ \sin^{-1}\dfrac{3}{5}=x.$$\\$ Then,$ \sin x=\dfrac{3}{5}\\ \implies \cos x=\sqrt{1-\sin^2 x} =\dfrac{4}{5}\\ \implies \sec x=\dfrac{5}{4}\\ \therefore \tan x =\sqrt{\sec^2 x-1} =\sqrt{\dfrac{25}{16}-1} =\dfrac{3}{4}\\ \therefore x = \tan^{-1}\dfrac{3}{4}\\ \therefore \sin^{-1}\dfrac{3}{5} =\tan^{-1}\dfrac{3}{4}..................(i)$$\\$ Now , $ \cot^{-1}\dfrac{3}{2} =\tan^{-1}\dfrac{2}{3}..................(ii) \ \ \ \ \ \ \ \ \ \ [\therefore = \cot^{-1}x]$ $\\$ Therefore , $ \tan(\sin^{-1}\dfrac{3}{5}+\cot^{-1}\dfrac{3}{2})\\ =\tan(\tan^{-1}\dfrac{3}{4}+ \tan^{-1}\dfrac{2}{3}) \ \ \ \ \ \ \ \ \ [\text{Using (i) and (ii)}]\\ =\tan \left[\tan^{-1}(\dfrac{\dfrac{3}{4}+\dfrac{2}{3}}{1-\dfrac{3}{4}.\dfrac{2}{3}}) \right] \ \ \ \ \ \ \ [As,\tan^{-1} x+ \tan^{-1} y = \tan^{-1} \dfrac{x+y}{1-xy}]\\ = \tan(\tan^{-1}\dfrac{9+8}{12-6})\\ = \tan(\tan^{-1}\dfrac{17}{6}= \dfrac{17}{6})$

34   Find the values of $ \cos^{-1} (\cos \dfrac{7\pi}{6})$ is equal to

Solution :

35   Find the values of $ \cos^{-1} (\cos \dfrac{7\pi}{6})$ is equal to

Solution :

We know that $\cos^{-1} (\cos x) =x$ if $ x \in [0,\pi],$ which is the principal value branch of $\cos^{-1}x.$$\\$ Here,$\dfrac{7\pi}{6}\notin [0,\pi].$$\\$ Now,$ \cos^{-1}(\cos \dfrac{7\pi}{6})$ can be written as:$\\$ $\cos^{-1}(\cos\dfrac{7\pi}{6}) =\cos^{-1}(\cos\dfrac{-7\pi}{6}) \\ =\cos^{-1}[\cos(2\pi-\dfrac{7\pi}{6})] \ \ \ \ \ \ \ [\therefore +x)=\cos x]\\ \therefore \cos^{-1}(\cos \dfrac{7\pi}{6})\\ =\cos^{-1}(\cos\dfrac{5\pi}{6}) =\dfrac{5\pi}{6}$$\\$ The correct answer is B.

36   Find the value of $ \sin(\dfrac{\pi}{3}-\sin^{-1}(-\dfrac{1}{2}))$ is equal to

Solution :

Let $ \sin^{-1}(\dfrac{-1}{2})=x$$\\$ Then, $ \sin x =\dfrac{-1}{2}=-\sin \dfrac{\pi}{6} =\sin(\dfrac{-\pi}{6}).$ $\\$ We know that the range of the principal value branch of $ \sin^{-1}$ is $[\dfrac{-\pi}{2},\dfrac{\pi}{2}].$$\\$ $\sin^{-1}(\dfrac{-1}{2})=\dfrac{\pi}{6}\\ \therefore \sin(\dfrac{\pi}{3}-\sin^{-1}(\dfrac{-1}{2})) =\sin(\dfrac{\pi}{3}+\dfrac{\pi}{6})=\sin(\dfrac{3\pi}{6})=\sin(\dfrac{\pi}{2}) =1$ The correct answer is D.

37   Find the value of $\cos^{-1}(\cos \dfrac{13\pi}{6})$

Solution :

We Know that $\cos^{-1}(\cos x) =x $ if $ x\in [0,\pi],$ which is the principal value branch of $ \cos^{-1}x.$$\\$ Here , $ \dfrac{13 \pi}{6}\notin [0,\pi].$$\\$ Now, $ \cos^{-1}(\cos \dfrac{13\pi}{6})$ can be written as:$\\$ $\cos^{-1}(\cos\dfrac{13\pi}{6})\\ =\cos^{-1} [\cos(2\pi+\dfrac{\pi}{6})]\\ =\cos^{-1}[\cos(\dfrac{\pi}{6})] $ where $ \dfrac{\pi}{6}\in [0,\pi].$ $\\$ $\therefore \cos^{-1} (\cos\dfrac{13\pi}{6}) =\cos^{-1}[\cos(\dfrac{\pi}{6})] =\dfrac{\pi}{6}$

38   Find the value of $\tan^{-1}(\tan \dfrac{7\pi}{6})$

Solution :

We know that $\tan^{-1} (\tan x) =x $ if $ x\in (-\dfrac{\pi}{2},\dfrac{\pi}{2}),$ which is the principal value branch of $ \tan^{-1} x.$ $\\$ Here,$ \dfrac{7\pi}{6}\notin (-\dfrac{\pi}{2},\dfrac{\pi}{2}).$$\\$ Now, $ \tan^{-1}(\tan \dfrac{7\pi}{6})$ can be written as:$\\$ $\tan^{-1}(\tan \dfrac{7\pi}{6})\\ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ -x)=-\tan x]\\ =\tan^{-1}[\tan(2\pi -\dfrac{5\pi}{6})] \\ =\tan^{-1}[-\tan(\dfrac{5\pi}{6})] \ \ \ \ \ [\therefore \ \ \ \ -x)=-\tan x] \\ =\tan^{-1}[\tan (\pi-\dfrac{5\pi}{6})]\\ =\tan^{-1}[\tan(\dfrac{\pi}{6})],$ where $ \dfrac{\pi}{6}\in(-\dfrac{\pi}{2},\dfrac{\pi}{2})$$\\$ $\therefore \tan^{-1}(\tan\dfrac{7\pi}{6}) =\tan^{-1}(\tan \dfrac{\pi}{6})=\dfrac{\pi}{6}$

39   Prove $ 2 \sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}$

Solution :

Let $ \sin^{-1} \dfrac{3}{5}=x.$ Then $\sin x=\dfrac{3}{5}.$$\\$ $\implies \cos x=\sqrt{1-(\dfrac{3}{5})^2} =\dfrac{4}{5}\\ \therefore \tan x=\dfrac{3}{4}\\ \therefore x=\tan^{-1}\dfrac{3}{4} \implies \sin^{-1}\dfrac{3}{5} =\tan^{-1}\dfrac{3}{4}$$\\$ Now , we have: $\\$ L.H.S $\\$ $ 2\sin^{-1}\dfrac{3}{5}=2\tan^{-1}\dfrac{3}{4}\\ =\tan^{-1}(\dfrac{2*dfrac{3}{4}}{1-(\dfrac{3}{4})^2}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ \ \ \ \ \ \ \ \ =\tan^{-1}\dfrac{2x}{1-x^2}]\\ =\tan^{-1}(\dfrac{dfrac{3}{2}}{\dfrac{16-9}{16}}) \\ =\tan^{-1} (\dfrac{3}{2}*\dfrac{16}{7})\\ =\tan^{-1} \dfrac{24}{7} = R.H.S.$

40   Prove $\sin^{-1}\dfrac{8}{17}+ \sin^{-1}\dfrac{3}{5} =\tan^{-1} \dfrac{77}{36}$

Solution :

$\sin ^{-1}\dfrac{8}{17} =x.$$\\$ Then ,$ \sin x =\dfrac{8}{17} \implies \cos x =\\ \sqrt{1-(\dfrac{8}{17})^2} =\sqrt{\dfrac{225}{289}} =\dfrac{15}{17}. \\ \therefore \tan x=\dfrac{8}{15} \implies x=\tan^{-1}\dfrac{8}{15}\\ \therefore \sin^{-1} \dfrac{8}{17} = \tan^{-1} \dfrac{8}{15} ...........(1)$$\\$ Now, let $ \sin^{-1} \dfrac{3}{5} = y $$\\$ Then , $ \sin y =\dfrac{3}{5} \implies \cos y =\sqrt{1-(\dfrac{3}{5})^2} =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5}.\\ \therefore \tan y =\dfrac{3}{5} \implies y=\tan^{-1}\dfrac{3}{4}\\ \therefore \sin^{-1} \dfrac{3}{5} =\tan^{-1}\dfrac{3}{4} ..............(2)$ Now , we have: $\\$ L.H.S. $\\$ $\sin^{-1}\dfrac{8}{17}+ \sin^{-1} \dfrac{3}{5}$$\\$ Using (1) and (2), we get $\\$ $=\tan^{-1} \dfrac{8}{15} +\tan^{-1} \dfrac{3}{4} \\ =\tan^{-1}\dfrac{\dfrac{8}{15}+\dfrac{3}{4}}{1-\dfrac{8}{15}*\dfrac{3}{4}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ \ \ \ \ \vdash \tan^{-1} y =\tan^{-1}\dfrac{x+y}{1-xy}] \\ = \tan^{-1}(\dfrac{32+45}{60-24}) \\ =\tan^{-1}\dfrac{77}{36} =R.H.S.$

41   Prove $\cos^{-1} \dfrac{4}{5} + \cos^{-1}\dfrac{12}{13} =\cos^{-1}\dfrac{33}{65}$

Solution :

Let $\cos^{-1} \dfrac{4}{5} =x$$\\$ Then,$ \cos x=\dfrac{4}{5} \implies \sin x=\sqrt{1-(\dfrac{4}{5})^2} =\dfrac{3}{5} \\ \therefore \tan x=\dfrac{3}{4} \implies x=\tan^{-1}\dfrac{3}{4} \\ \therefore \cos^{-1}\dfrac{4}{5} =\tan^{-1}\dfrac{3}{4} .........(1)$$\\$ Now, let$ \cos^{-1}\dfrac{12}{13} =y.$$\\$ Then,$\cos y =\dfrac{12}{13} \implies \sin y =\dfrac{5}{13}.\\ \therefore \tan y =\dfrac{5}{12} \implies y = \tan^{-1}\dfrac{5}{12}\\ \therefore \cos^{-1}\dfrac{12}{13} =\tan^{-1}\dfrac{5}{12} .........(2)$$\\$ Let$\cos^{-1} \dfrac{33}{65} =z$$\\$ Then , $ \cos z = \dfrac{33}{65} \implies \sin z =\dfrac{56}{65}\\ \therefore \tan z=\dfrac{56}{33} \implies z=\tan^{-1}\dfrac{56}{33}\\ \therefore \cos^{-1} \dfrac{33}{65} =\tan^{-1}\dfrac{56}{33} .........(3)$ $\\$ Now , L.H.S. $\\$ $\cos^{-1}\dfrac{4}{5} +\cos^{-1}\dfrac{12}{13}$$\\$ Using (1) and (2),we get $\\$ $=\tan^{-1}\dfrac{3}{4} + \tan^{-1} \dfrac{5}{12}\\ =\tan^{-1}(\dfrac{dfrac{3}{4}+\dfrac{5}{12}}{1-(\dfrac{3}{4}*\dfrac{5}{12})}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ \ \vdash \tan^{-1} y= \tan^{-1}(\dfrac{x+y}{1-xy})]\\ =\tan^{-1} (\dfrac{36+20}{48-15})\\ =\tan^{-1}\dfrac{56}{33}\\ =\cos^{-1} \dfrac{56}{33}R.H.S $

42   Prove $\cos^{-1} \dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}$

Solution :

Let $ \sin^{-1}\dfrac{3}{5} =x $ Then, $ \sin x=\frac{3}{5} \implies \cos x=\sqrt{1-(\dfrac{3}{5})^2} =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5}\\ \therefore \tan x=\dfrac{3}{4} \implies x=\tan^{-1}\dfrac{3}{4} \\ \therefore \sin^{-1} \dfrac{3}{5} =\tan^{-1}\dfrac{3}{4}..........(1)$$\\$ Now,let $ \cos^{-1}\dfrac{12}{13} =y$$\\$ Then $ \cos y =\dfrac{12}{13} \implies \sin y =\dfrac{5}{13}.\\ \therefore \tan y =\dfrac{5}{12} \implies y=\tan^{-1}\dfrac{5}{12}\\ \therefore \cos^{-1}\dfrac{12}{13} =\tan^{-1}\dfrac{5}{12} ...........(2)$$\\$ Let $\sin^{-1} \dfrac{56}{65} = z.$$\\$ Then $,\sin z =\dfrac{56}{65} \implies \cos z=\dfrac{33}{65}\\ \therefore \tan z=\dfrac{56}{33} \implies z =\tan^{-1}\dfrac{56}{33}\\ \therefore \sin^{-1} \dfrac{56}{65} = \tan^{-1} \dfrac{56}{33}......(3)$$\\$ Now,we have $\\$ L.H.S. $\\$ $\cos^{-1}\dfrac{12}{13}+ \sin^{-1}\dfrac{3}{5} \\ =\tan^{-1} \dfrac{5}{12} + \tan^{-1}\dfrac{3}{4}$$\\$ Using (1)and (2), we get$\\$ $=\tan^{-1}\dfrac{\dfrac{5}{12}+\dfrac{3}{4}}{1-(\dfrac{5}{12}.\dfrac{3}{4})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ \vdash \tan^{-1}y =\tan^{-1}\dfrac{x+y}{1-xy}]\\ =\tan^{-1}(\dfrac{20+36}{48-15})\\ =\tan^{-1} \dfrac{56}{33}\\ = \sin^{-1} \dfrac{56}{65} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [Using (3)] =R.H.S.$

43   Prove $ \tan^{-1}\dfrac{63}{16} = \sin^{-1} \dfrac{5}{13}+ \cos^{-1}\dfrac{3}{5}$

Solution :

Let $ \sin^{-1}\dfrac{5}{13} =x$$\\$ Then,$ \sin x=\dfrac{5}{13} \implies \cos x =\dfrac{12}{13}.\\ \therefore \tan x=\dfrac{5}{12} \implies x=\tan^{-1}dfrac{5}{12}\\ \therefore \sin^{-1} \dfrac{5}{13} =\tan^{-1}\dfrac{5}{12} ........(1)$$\\$ Let $ \cos^{-1} \dfrac{3}{5} =y.$$\\$ Then$ \cos y =\dfrac{3}{5} \implies \sin y =\dfrac{4}{5}.$$\\$ Thus, $ \tan y =\dfrac{4}{3} \implies y=\tan^{-1}\dfrac{4}{3}\\ \therefore \cos^{-1}\dfrac{3}{5} =\tan^{-1}\dfrac{4}{3} .........(2)$$\\$ Using (1) and (2) , we have $\\$ R.H.S. $\\$ $\sin^{-1}\dfrac{5}{12 }+ \cos^{-1}\dfrac{3}{5}\\ =\tan^{-1}(\dfrac{\dfrac{5}{12}+\dfrac{4}{3}}{1-\dfrac{5}{12}*\dfrac{4}{3}}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\therefore \ \ \ \ \ \ \vdash \tan^{-1}y = \tan^{-1}\dfrac{x+y}{1-xy}]\\ =\tan^{-1}(\dfrac{15+ 48}{36-20}) \\ =\tan^{-1} \dfrac{63}{16}= L.H.S. $

44   Prove $\tan^{-1}\dfrac{1}{5} + \tan^{-1}\dfrac{1}{7} + \tan^{-1} \dfrac{1}{3}+\tan^{-1}\dfrac{1}{8} =\dfrac{\pi}{4}$$\\$

Solution :

L.H.S. $\tan^{-1} \dfrac{1}{5}+ \tan^{-1}\dfrac{1}{7}+ \tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{8}$$ \\$ $=\tan^{-1}(\dfrac{\dfrac{1}{5}+\dfrac{1}{7}}{1-\dfrac{1}{5}*\dfrac{1}{7}})+ \tan^{-1}(\dfrac{dfrac{1}{3}+\dfrac{1}{8}}{1-\dfrac{1}{3}*\dfrac{1}{8}})$ $ \ \ \ \ \ \ \ \ \ \ [\tan^{-1}x + \tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}]$$\\$ $=\tan^{-1} \dfrac{(7+5)}{(35-1)} + \tan^{-1}(\dfrac{8+3}{24-1})$$\\$ $=\tan^{-1}\dfrac{12}{34}+\tan^{-1}\dfrac{11}{23}$$\\$ $=\tan^{-1}\dfrac{6}{17} +\tan^{-1}\dfrac{11}{23}\\ =\tan^{-1} (\dfrac{\dfrac{6}{17}+\dfrac{11}{23}}{1-\dfrac{6}{17}*\dfrac{11}{23}})\\ =\tan^{-1}(\dfrac{325}{325}) =\tan^{-1} 1\\ =\dfrac{\pi}{4} =R.H.S.$

45   Prove $\tan^{-1}\sqrt{x} =\dfrac{1}{2}\cos^{-1} (\dfrac{1-x}{1+x}),x\in [0.1]$

Solution :

Let $ x=\tan^2 \theta . $$\\$ Then $\sqrt{x}=\tan \theta \implies \theta =\tan^{-1} \sqrt{x}.$$\\$ $ \therefore \dfrac{1-x}{1+x} =\dfrac{1-\tan^2 \theta }{1+ \tan^2 \theta } =\cos 2\theta $$\\$ Now, we have: $\\$ R.H.S. $ \dfrac{1}{2}\cos^{-1}(\dfrac{1-x}{1+x})\\ =\dfrac{1}{2}\cos^{-1}(\cos 2 \theta) \\ =\dfrac{1}{2} * 2 \theta =\theta =\tan^{-1} \sqrt{x} = L.H.S. $

46   Prove $ \cot^{-1}(\dfrac{\sqrt{1+ \sin x}+\sqrt{1-\sin x}}{\sqrt{1+ \sin x}-\sqrt{1- \sin x}}) = \dfrac{x}{2},x \in (0,\dfrac{\pi}{4})$

Solution :

Consider $ \dfrac{\sqrt{1+ \sin x}+ \sqrt{1-\sin x}}{\sqrt{1+\sin x-\sqrt{1-\sin x}}}\\ =\dfrac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})^2} \ \ \ \ \ \ \ \ \ \ \ (by rationalizing)\\ =\dfrac{(1+\sin x)+(1-\sin x)+ 2\sqrt{(1+\sin x)(1- \sin x)}}{1+ \sin x - 1+ \sin x} \\ =\dfrac{2(1+\sqrt{1-\sin^2 x})}{2 \sin x} \\ =\dfrac{1+ \cos x}{\sin x}\\ = \dfrac{2\cos^2 \dfrac{x}{2}}{2 \sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\ =\cot \dfrac{x}{2} \\ \therefore L.H.S. =\cot^{-1}(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}})\\ \cot^{-1}(\cot \dfrac{x}{2}) =\dfrac{x}{2} = R.H.S.$

47   Prove $\tan^{-1}(\dfrac{\sqrt{1+ x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}) =\dfrac{\pi}{4} -\dfrac{1}{2} \cos^{-1} x, -\dfrac{1}{\sqrt{2}} \leq x \leq 1 $ $\\$ [Hint : put x= $ \cos 2 \theta $]

Solution :

Let, $x =\cos 2 \theta $ then, $\theta =\dfrac{1}{2} \cos^{-1} x$$\\$ Thus , we have:$\\$ $ L.H.S. = \tan^{-1}(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}})\\ =\tan^{-1}(\dfrac{\sqrt{1+\cos 2 \theta} -\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}) \\ =\tan^{-1}(\dfrac{\sqrt{2 \cos^2 2 \theta }-\sqrt{2 \sin^2 \theta }}{\sqrt{2 \cos^2 2 \theta }+\sqrt{2 \sin^2 \theta}})\\ =\tan^{-1}(\dfrac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta }{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta })\\ =\tan^{-1}(\dfrac{\cos \theta -\sin \theta }{\cos \theta + \sin \theta })\\ =\tan^{-1}(\dfrac{1-\tan \theta }{1+ \tan \theta })\\ =\tan ^{-1}1 - \tan^{-1}(\tan \theta ) \ \ \ \ \ \ [\therefore (\dfrac {x-y}{1+ xy}) =\tan^{-1} x-\tan^{-1} y]\\ =\dfrac{\pi}{4}-\theta =\dfrac{\pi}{4} -\dfrac{1}{2}\cos^{-1} x =R.H.S.$

48   $ Prove.=\dfrac{9\pi}{8}-\dfrac{9}{4} \sin^{-1} \dfrac{1}{3} =\dfrac{9}{4}\sin ^{-1} \dfrac{2\sqrt{2}}{3}$

Solution :

$ L.H.S.=\dfrac{9\pi}{8}-\dfrac{9}{4} \sin^{-1} \dfrac{1}{3}\\ =\dfrac{9}{4}(\dfrac{\pi}{2}-\sin^{-1}\dfrac{1}{3}) \\ \dfrac{9}{4}(\cos ^{-1} \dfrac{1}{3}) \\ =\dfrac{9}{4}(\cos^{-1}\dfrac{1}{3}) .......(1) [\therefore -\cos^{-1} x =\dfrac{\pi}{2}] $$\\$ Now , let $ \cos^{-1} \dfrac{1}{3}=x$$\\$ Then,$ \cos x=\dfrac{1}{3} \implies \sin x =\sqrt{1-(\dfrac{1}{3})^2} =\dfrac{2 \sqrt{2}}{3}.\\ \therefore x= \sin^{-1} \dfrac{2\sqrt{2}}{3} \implies \cos^{-1}\dfrac{1}{3} \implies \sin^{-1} \dfrac{2\sqrt{2}}{3} \\ \therefore L.H.S. = \dfrac{9}{4}\sin^{-1} \dfrac{2\sqrt{2}}{3} =R.H.S.$

49   Solve $2 \tan^{-1}(\cos x) =\tan^{-1}(2 \csc x)$

Solution :

$2 \tan^{-1}(\cos x) =\tan^{-1}(2 \csc x)$$\\$ $\implies \tan^{-1}(\dfrac{2\cos x}{1-\cos^2 x}) =\tan^{-1} (2 \csc x) \ \ \ \ \ \ \ \ \ [\therefore = \tan^{-1} \dfrac{2x}{1-x^2}]$

$\implies \dfrac{2 \cos x}{\sin^2 x} =\dfrac{2}{\sin x}\\ \implies \cos x = \sin x\\ \implies \tan x =1 \\ \therefore x=\dfrac{\pi}{4}$

50   Solve $ \tan^{-1} \dfrac{1-x}{1+x}=\dfrac{1}{2} \tan^{-1}x,(x > 0)$

Solution :

$ \tan^{-1} \dfrac{1-x}{1+x}=\dfrac{1}{2} \tan^{-1}x\\ \Rightarrow \tan^{-1}1-\tan^{-1}x=\dfrac{1}{2}\tan^{-1}x\qquad [ \therefore -\tan^{-1}y=\tan^{-1}\dfrac{x-y}{1+xy}]\\ \Rightarrow \dfrac{\pi}{4}=\dfrac{3}{2}\tan^{-1}x\\ \Rightarrow \tan^{-1} x=\tan{\pi}{6}\\ \Rightarrow x=\tan \dfrac{\pi}{6}\\ \therefore x=\dfrac{1}{\sqrt{3}}$

51   Solve $\sin(\tan^{-1} x),|x| < 1 $ is equal to

Solution :

$\tan y=x \Rightarrow \sin y=\dfrac{x}{\sqrt{1+x^2}}$$\\$ Let $ \tan^-1x=y.\text{Then}\\ y=\sin^{-1}(\dfrac{x}{\sqrt{1+x^2}})\\ \Rightarrow \tan^{-1} x = \sin^{-1}(\dfrac{x}{\sqrt{1+x^2}})\\ \Rightarrow \sin(\tan^{-1}x)=\sin(\sin^{-1}(\dfrac{x}{\sqrt{1+x^2}}))=\dfrac{x}{1+x^2}$$\\$ The correct answer is D.

52   Solve $ \sin^{-1}(1-x)-2 \sin^{-1}x=\dfrac{\pi}{2}$,then $x$ is equal to

Solution :

$\sin^{-1}(1-x)-2 \sin^{-1}x=\dfrac{\pi}{2}\\ \Rightarrow -2\sin^{-1} x=\dfrac{\pi}{2}-\sin^{-1}(1-x)\\ \Rightarrow -2 \sin^{-1}x=\cos^{-1}(1-x) \quad ....(1)\\ \text{Let} \sin^{-1}x=\theta \Rightarrow \sin \theta =x\\ \Rightarrow \cos \theta =\sqrt{1-x^2}\\ \therefore \theta =\cos^{-1}(\sqrt{1-x^2})\\ \therefore \sin^{-1} x=\cos^{-1}(\sqrt{1-x^2})$$\\$ Therefore , from equation (1) , we have $\\$ $-2 \cos^{-1}(\sqrt{1-x^2})=\cos^{-1}(1-x)$$\\$ Let,$x=\sin y $. Then , we have:$\\$ $-2 \cos^{-1}(\sqrt{1-\sin^2 y}) = \cos^{-1}(1-\sin y)\\ \Rightarrow -2 \cos^{-1}(\cos y)=\cos ^{-1}(1-\sin y)\\ \Rightarrow -2y=\cos ^{-1}(1-\sin y)\\ \Rightarrow 1-\sin y =\cos (-2y)=\cos2y\\ \Rightarrow 1-\sin y=1-2 \sin^2 y\\ \Rightarrow 2\sin^2 y-\sin y =0\\ \Rightarrow \sin y(2 \sin y-1)=0\\ \Rightarrow \sin y =0 or \dfrac{1}{2}\\ \therefore x=0 \text{Or} x=\dfrac{1}{2}$$\\$ When $ x=\dfrac{1}{2},$ it can be observed that:$\\$ $L.H.S =\sin^{-1}(1-\dfrac{1}{2})-\sin^{-1}\dfrac{1}{2}\\ =\sin^{-1}(\dfrac{1}{2}) -2 \sin^{-1}\dfrac{1}{2}\\ =-\sin^{-1}\dfrac{1}{2}\\ =\dfrac{\pi}{6 }\neq \dfrac{\pi}{2} \neq R.H .S\\ \therefore x=\dfrac{1}{2}$ is not a solution of the given equation.$\\$ Thus , x=0. Hence, the correct answer is C

53   Solve $\tan^{-1}(\dfrac{x}{y})-\tan^{-1} \dfrac{x-y}{x+y}$ is equal to

Solution :

$\tan^{-1}(\dfrac{x}{y})-\tan ^{-1}(\dfrac{x-y}{x+y})\\ =\tan^{-1}\left[\dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+(\dfrac{x}{y})(\dfrac{x-y}{x+y})}\right] \quad [\therefore -\tan^{-1}y =\tan^{-1}\dfrac{x-y}{1+xy}]\\ =\tan^{-1}\left[\dfrac{\dfrac{x(x+y)-y(x-y)}{y(x+y)}}{\dfrac{y(x+y)+x(x-y)}{y(x+y)}}\right]\\ =\tan^{-1}(\dfrac{x^2+xy-xy+y^2}{xy+y^2+x^2-xy})\\ =\tan^{-1}(\dfrac{x^2+y^2}{x^2+y^2})=\tan^{-1}1=\dfrac{\pi}{4}$$\\$ Hence, the correct answer is C .