Solution:

Option(B) is correct.
To Find: The value of $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
Let, $x=\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
$\begin{array}{l}
\Rightarrow \mathrm{x}=\tan ^{-1}\left\{2 \cos \left(2\left(\frac{\pi}{6}\right)\right)\right\}\left(\because \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}\right) \\
\Rightarrow \mathrm{x}=\tan ^{-1}\left(2 \cos \frac{\pi}{3}\right) \\
\Rightarrow \mathrm{x}=\tan ^{-1}\left(2\left(\frac{1}{2}\right)\right)=\tan ^{-1} 1=\frac{\pi}{4}\left(\because \cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \text { and } \tan \left(\frac{\pi}{4}\right)=1\right)
\end{array}$