Find the value of

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

$\therefore \sin ^{-1} \frac{1}{2}=\frac{\pi}{6}$

$\therefore \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$

$=\tan ^{-1}\left[2 \cos \frac{\pi}{3}\right]=\tan ^{-1}\left[2 \times \frac{1}{2}\right]$

$=\tan ^{-1} 1=\frac{\pi}{4}$