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 $\sin ^{-1} x + \sin ^{-1} y = \frac{\pi}{2}$,then $x^{2}$ is equal to

The value of $\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$,where $x \in\left(0, \frac{\pi}{4}\right)$ is

$\tan \left[ {\frac{\pi }{4} + \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] + \tan \left[ {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] = $

$\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$,then $x=$ . . . . . . .

$ \cos \left[2 \sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4}\right] $

Vedclass Test Series

Exam Paper Generator

Online Exam Module