કિંમત શોધો: sin ^{-1}left(frac{3}{5}right) + | vedclass

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(A) ધારો કે $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. તેથી $\sin 	heta = \frac{3}{5}$.નિત્યસમ $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin ^

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$\frac{\pi }{4}$

Vedclass · Answer

(A) ધારો કે $	heta = \sin ^{-1}\left(\frac{3}{5}ight)$. તેથી $\sin 	heta = \frac{3}{5}$.નિત્યસમ $	an 	heta = \frac{\sin 	heta}{\sqrt{1 - \sin ^2 	heta}} = \frac{3/5}{\sqrt{1 - (3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4}$ નો ઉપયોગ કરતા.આમ,$\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.હવે,પદાવલિ $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{7}ight)$ બને છે.સૂત્ર $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ ($xy $	an ^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} 	imes \frac{1}{7}}ight) = 	an ^{-1}\left(\frac{\frac{21+4}{28}}{1 - \frac{3}{28}}ight) = 	an ^{-1}\left(\frac{25/28}{25/28}ight) = 	an ^{-1}(1) = \frac{\pi }{4}$.

કિંમત શોધો: $\sin ^{-1}\left(\frac{3}{5}\right) + \tan ^{-1}\left(\frac{1}{7}\right) = $

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