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

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(A) આપણે જાણીએ છીએ કે $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$ જ્યાં $|x| \le \frac{1}{\sqrt{2}}$.અહીં,$x = \frac{3}{5}$ છે,તેથી $2{\

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$ જ્યાં $|x| \le \frac{1}{\sqrt{2}}$.અહીં,$x = \frac{3}{5}$ છે,તેથી $2{\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\left( 2 	imes \frac{3}{5} 	imes \sqrt {1 - \frac{9}{{25}}} ight) = {\sin ^{ - 1}}\left( \frac{6}{5} 	imes \frac{4}{5} ight) = {\sin ^{ - 1}}\frac{{24}}{{25}}$.હવે,પદાવલિ ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}}$ બને છે.નિત્યસમ ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$ નો ઉપયોગ કરતા,આપણને ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}} = \frac{\pi }{2}$ મળે છે.

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

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