Find the value of tan ^{-1}left[2 cos left(2 | vedclass

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(B) Let $\sin ^{-1} \left(\frac{1}{2}\right) = x$. Then,$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$. Therefore,$\sin ^{-1} \left(\frac{1}

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $\sin ^{-1} \left(\frac{1}{2}ight) = x$. Then,$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}ight)$. Therefore,$\sin ^{-1} \left(\frac{1}{2}ight) = \frac{\pi}{6}$. Now,substitute this value into the expression: $	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]$. $= 	an ^{-1} \left[2 \cos \left(\frac{\pi}{3}ight)ight]$. Since $\cos \left(\frac{\pi}{3}ight) = \frac{1}{2}$,we have: $= 	an ^{-1} \left[2 	imes \frac{1}{2}ight] = 	an ^{-1} (1)$. Since $	an \left(\frac{\pi}{4}ight) = 1$,the value is $\frac{\pi}{4}$.

Find the value of $\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$

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