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

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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}}

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$\frac{\pi }{2}$

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}} = $

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