# Inverse Trigonometric Functions

## Class 12 NCERT

### NCERT

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

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 $\csc ^{-1} (2)$

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 $\tan^{-1}(-1)$

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 $\cos ^{-1} (-\dfrac{1}{\sqrt{2}})$

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

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:$\tan^{-1}(\dfrac{1}{\sqrt{x^2-1}}),|x|>1$

$\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:$\tan^{-1}(\dfrac{3a^2 x- x^3}{a^3 - 3ax^2}),a > 0 ; \dfrac{-a}{\sqrt{3}} \leq x \leq \dfrac{a}{\sqrt{3}}$

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 $xy < 1$

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.

$\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

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= $\cos 2 \theta$]

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