Given $\tan^{-1}\begin{bmatrix}2\sin\bigg(2\cos^{-1}\frac{\sqrt 3}{2}\bigg)\end{bmatrix}$
$\textbf{Step 1:} \; 2 \;\mathbf{\cos^{-1}\frac{\sqrt 3}{2}}$
The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Let $\cos^{-1}(\large\frac{\sqrt 3}{2})$$ = x$ $ \Rightarrow \cos x = \large\frac{\sqrt 3}{2}$
Therefore, $\cos x = \large\frac{\sqrt 3}{2} = \cos \large\frac{\pi}{6}$
$\Rightarrow x = \large\frac{\pi}{6}$, where $x \;\epsilon \; \left [ 0,\pi \right ]$
$\Rightarrow 2 \mathbf{\cos^{-1}\frac{\sqrt 3}{2}} = 2\large \frac{\pi}{6} $$= \large \frac {\pi}{3}$
The expression reduces to $\tan^{-1}(2\sin\frac{\pi}{3})$
$\textbf{Step 2:} \; 2\; \mathbf{\sin\frac{\pi}{3}}$
We know that $\sin\frac{\pi}{3} = \frac{\sqrt3}{2}$
$\Rightarrow 2\; \mathbf{\sin\frac{\pi}{3}} = 2\frac{\sqrt3}{2} = \sqrt (3)$
The expression reduces to $\tan^{-1} \sqrt (3)$
$\textbf{Step 3:} \;tan^{-1} \sqrt (3)$
We know that $\tan^{-1}\sqrt (3) = \frac{\pi}{3}$
Therefore, the expression $\tan^{-1}\begin{bmatrix}2\sin\bigg(2\cos^{-1}\frac{\sqrt 3}{2}\bigg)\end{bmatrix}$ reduces to $\frac{\pi}{3}$