Find the value of: $\;$ $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right)$
$\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{3}\right)\right] = \dfrac{\pi}{3}, \hspace{1cm} \dfrac{\pi}{3} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$\cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) = \pi - \cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}, \hspace{1cm} \dfrac{5 \pi}{6} \in \left[0, \pi\right]$
$\tan^{-1} \left(1\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
$\begin{aligned}
\therefore \; \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right) & = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + 2 \times \dfrac{\pi}{4} \\\\
& = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + \dfrac{\pi}{2} \\\\
& = \dfrac{2 \pi - 5 \pi + 3 \pi}{6} \\\\
& = 0
\end{aligned}$