EXERCISE 2.2

Prove the following:(Class 12 ncert solution math exercise 2.2)

Question 1: Prove $3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^{3}\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$ .

Solution: Let $x=\sin \theta x=\sin \theta$

Hence, $\sin ^{-1}(x)=\theta \sin ^{-1}(x)=\theta$

Now,

$RHS =\sin ^{-1}\left(3 x-4 x^{3}\right)$

$=\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)$

$=\sin ^{-1}(\sin 3 \theta)$

$=3 \theta$

$=3 \sin ^{-1} x$

=L H S

Question 2: Prove $3 \cos ^{-1} x=\cos ^{-1}\left(4 x^{3}-3 x\right), x \in\left[\frac{1}{2}, 1\right]$ .

Solution: Let $x=\cos \theta$

Hence, $\cos ^{-1}(x)=\theta$

Now,

$RHS =\cos ^{-1}\left(4 x^{3}-3 x\right)$

$=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right)$

$=\cos ^{-1}(\cos 3 \theta)$

$=3 \theta$

$=3 \cos ^{-1} x$

$=L H S$

Question 3: Prove $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$ .

Solution: Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$

Now,

$L H S =\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}$

$=\tan ^{-1} \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \cdot \frac{7}{24}}$

$=\tan ^{-1}\left(\frac{\frac{48+77}{264}}{\frac{264-14}{264}}\right)$

$=\tan ^{-1}\left(\frac{125}{250}\right)$

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

$=R H S$

Question 4:Prove $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$ .

Solution: Since we know that

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

$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$
Now,

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

$=\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$

$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}\right)$

$=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{21-4}\right)$

$=\tan ^{-1}\left(\frac{31}{17}\right)$

$=R H S$

Question 5: Write the function in the simplest form: $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$

Solution: Let $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

Hence,

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

$=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$

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

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

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

$=\frac{\theta}{2}$

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

Question 6: Write the function in the simplest form: $\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}},|x|>1$

Solution: Let $x=\operatorname{cosec} \theta \Rightarrow \theta=\operatorname{cosec}^{-1} x$

Hence,

$\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}} =\tan ^{-1} \frac{1}{\sqrt{\operatorname{cosec}^{2} \theta-1}}$

$=\tan ^{-1}\left(\frac{1}{\cot \theta}\right)$

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

$=\theta$

$=\operatorname{cosec}{ }^{-1} x$

$=\frac{\pi}{2}-\sec ^{-1} x$

Question 7: Write the function in the simplest form: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0<x<\pi$

Solution: Since, $1-\cos x=2 \sin ^{2} \frac{x}{2}$

$1+\cos x=2 \cos ^{2} \frac{x}{2}$

Hence,

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

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

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

$=\frac{x}{2}$

Question 8: Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), 0<x<\pi$

Solution: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) =\tan ^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{\cos x}}\right)$

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

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

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

$=\frac{\pi}{4}-x$

Question 9: Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}},|x|<a$

Solution: Let $x=a \sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{a}\right)$

Hence,

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

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

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

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

$=\theta$

$=\sin ^{-1} \frac{x}{a}$

Question 10: Write the function in the simplest form: $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$

Solution: Let $x=a \tan \theta \Rightarrow \theta=\tan ^{-1}\left(\frac{x}{a}\right)$

Hence,

$\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right) =\tan ^{-1}\left(\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right)$

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

$=\tan ^{-1}(\tan 3 \theta)$

$=3 \theta$

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

Question 11: Write the function in the simplest form: $\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$

Solution: Let $\sin ^{-1} \frac{1}{2}=x$

Hence,

Therefore,

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

$=\sin \left(\frac{\pi}{6}\right)$

$x =\frac{\pi}{6}$

$\sin ^{-1}\left(\frac{1}{2}\right) =\frac{\pi}{6}$

$\tan -1\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right] =\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$

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

$=\tan ^{-1}\left[2 \times \frac{1}{2}\right]$

$=\tan ^{-1}[1]$

$=\frac{\pi}{4}$

Question 12: Find the value of $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$

Solution: Since ${ }^{\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}}$

Hence,

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

$=0$

Question 13: Find the value of $\tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right),|x|<1, y>0$ and $x y<1$ .

Solution: Let $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

Hence,

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

$=\sin ^{-1}(\sin 2 \theta)$

$=2 \theta$

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

Now, let $y=\tan \phi \Rightarrow \phi=\tan ^{-1} y$

Hence,

Therefore,

$\cos ^{-1} \frac{1-y^{2}}{1+y^{2}} =\cos ^{-1}\left(\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}\right)$

$=\cos ^{-1}(\cos 2 \phi)$

$=2 \phi$

$=2 \tan ^{-1} y$

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

$=\tan \left(\tan ^{-1} x+\tan ^{-1} y\right)$

$=\tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\left(\frac{x+y}{1-x y}\right)$

Question 14: If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$ , find the value of $x$ .

Solution: It is given that $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
$\Rightarrow \sin^{-1}\frac{1}{5}+\cos^{-1}x = \sin^{-1}1$

$\Rightarrow \sin^{-1}\frac{1}{5}+\cos^{-1}x = \frac{\pi}{2}$

$\Rightarrow \cos^{-1}x =\frac{\pi}{2}-\sin^{-1}\frac{1}{5}$

$\Rightarrow\cos^{-1}x= \cos^{-1}\frac{1}{5}$

$\Rightarrow x = \frac{1}{5}$

Question 15: If $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x-1}=\frac{\pi}{4}$ , find the value of $x$ .

Solution: It is given that $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x-1}=\frac{\pi}{4}$

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

Therefore,

$\Rightarrow \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x+2)(x-2)-(x-1)(x+1)}\right]=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left[\frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}\right]=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left[\frac{2 x^{2}-4}{-3}\right]=\frac{\pi}{4}$

$\Rightarrow \tan \left[\tan ^{-1} \frac{4-2 x^{2}}{3}\right]=\tan \frac{\pi}{4}$

$\Rightarrow \frac{4-2 x^{2}}{3}=1$

$\Rightarrow 4-2 x^{2}=3$

$\Rightarrow 2 x^{2}=1$

$\Rightarrow x^{2}=\frac{1}{2}$

$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$

Question 16: Find the value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ .

Solution: Since, $\sin \sin \quad(\not \infty-)$

Therefore,

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

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

$=\frac{\pi}{3}$

Question 17: Find the value of $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ .

Solution: $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) =\tan ^{-1}\left[-\tan \left(-\frac{3 \pi}{4}\right)\right]$

$=\tan ^{-1}\left[-\tan \left(\pi-\frac{\pi}{4}\right)\right]$

$=\tan ^{-1}\left[-\tan \left(\frac{\pi}{4}\right)\right]$

$=\tan ^{-1}\left[\tan \left(-\frac{\pi}{4}\right)\right]$

$=\left(-\frac{\pi}{4}\right)$

Question 18: Find the value of $\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$ .

Solution: Let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

Therefore,

$\Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5}$

$\Rightarrow \sec x=\frac{5}{4}$

Now,

$\tan x =\sqrt{\sec ^{2} x-1}$

$=\sqrt{\frac{25}{16}-1}$

$=\frac{3}{4}$

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

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

$\cot ^{-1} \frac{3}{2}=\tan ^{-1} \frac{2}{3}$

Thus, by using (1) and (2)

$\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{2}{3}\right) =\tan \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right)$

$=\tan \left[\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right]$

$=\tan \left(\tan ^{-1} \frac{17}{6}\right)$

$=\frac{17}{6}$

Question 19: $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal to

(A) $\frac{7 \pi}{6}$

(B) $\frac{5 \pi}{6}$

(D) $\frac{\pi}{6}$

Solution: $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) =\cos ^{-1}\left(\cos \frac{-7 \pi}{6}\right)$

$=\cos ^{-1}\left[\cos \left(2 \pi-\frac{7 \pi}{6}\right)\right]$

$=\cos ^{-1}\left[\cos \left(\frac{5 \pi}{6}\right)\right]$

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

Thus, the correct option is B.

Question 20: $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(D) 1

Solution: Let $\sin ^{-1}\left(-\frac{1}{2}\right)=x$

Hence,

$\sin x =-\frac{1}{2}$

$=-\sin \frac{\pi}{6}$

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

$x =-\frac{\pi}{6}$

Since, Range of principal value of $\sin ^{-1}(x)=\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ .

Therefore,

Then,

$\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$

$\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right) =\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$

$=\sin \left(\frac{\pi}{2}\right)$

Thus, the correct option is D.

Question 21: Find the values of $\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\frac{\pi}{2}$

(D) $2 \sqrt{3}$

Solution: Let $\tan ^{-1} \sqrt{3}=x$

Hence,

$\tan x=\sqrt{3}=\tan \frac{\pi}{3} \text {, where } \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Therefore, $\tan ^{-1} \sqrt{3}=\frac{\pi}{3}$

Now, let $\cot ^{-1}(-\sqrt{3})=y$

Hence,

$\cot y =(-\sqrt{3})$

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

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

$=\cot \left(\frac{5 \pi}{6}\right)$

Since, Range of principal value of $\cot ^{-1} x=(0, \pi)$

Therefore,

Then,

$\cot ^{-1}(-\sqrt{3})=\frac{5 \pi}{6}$

$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) =\frac{\pi}{3}-\frac{5 \pi}{6}$

$=-\frac{\pi}{2}$

Thus, the correct option is B.

EXERCISE 2.2

Leave a Comment Cancel reply