tan ^{-1}left[2 cos left(2 sin ^{-1} frac{1}{ | vedclass

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(B) ધારો કે $\sin ^{-1} \left(\frac{1}{2}\right) = x$. તેથી,$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$. તેથી,$\sin ^{-1} \left(\frac{1}{

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$\frac{\pi}{4}$

Vedclass · Answer

(B) ધારો કે $\sin ^{-1} \left(\frac{1}{2}ight) = x$. તેથી,$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}ight)$. તેથી,$\sin ^{-1} \left(\frac{1}{2}ight) = \frac{\pi}{6}$. હવે,આ કિંમતને પદાવલિમાં મૂકતા: $	an ^{-1} \left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}ight)ight] = 	an ^{-1} \left[2 \cos \left(2 	imes \frac{\pi}{6}ight)ight]$. $= 	an ^{-1} \left[2 \cos \left(\frac{\pi}{3}ight)ight]$. કારણ કે $\cos \left(\frac{\pi}{3}ight) = \frac{1}{2}$,તેથી: $= 	an ^{-1} \left[2 	imes \frac{1}{2}ight] = 	an ^{-1} (1)$. કારણ કે $	an \left(\frac{\pi}{4}ight) = 1$,તેથી જવાબ $\frac{\pi}{4}$ છે.

$\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$ ની કિંમત શોધો.

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