2{sin ^{ - 1}}frac{3}{5} + {cos ^{ - 1}}frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$ for $|x| \le \frac{1}{\sqrt{2}}$.Here,$x = \frac{3}{5}$,so $2{\sin ^{ - 1}}

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(A) We know that $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$ for $|x| \le \frac{1}{\sqrt{2}}$.Here,$x = \frac{3}{5}$,so $2{\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\left( 2 	imes \frac{3}{5} 	imes \sqrt {1 - \frac{9}{{25}}} ight) = {\sin ^{ - 1}}\left( \frac{6}{5} 	imes \frac{4}{5} ight) = {\sin ^{ - 1}}\frac{{24}}{{25}}$.Now,the expression becomes ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}}$.Using the identity ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}} = \frac{\pi }{2}$.

$2{\sin ^{ - 1}}\frac{3}{5} + {\cos ^{ - 1}}\frac{{24}}{{25}} = $

Similar Questions

If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$,then the value of $\sin x$ is

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ is equal to

If $\theta = \tan^{-1} a$,$\phi = \tan^{-1} b$ and $ab = -1$,then $\theta - \phi = $

Find the number of solutions for the equation $\sin^{-1} x = 2 \tan^{-1} x$ (in principal values).

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module