Find the value of: $\;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$
$\cot^{-1} \left(1\right) = \cot^{-1} \left[\cot \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(0, \pi\right)$
$\sin^{-1} \left(\dfrac{1}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$\csc^{-1} \left(- 2\right) = \csc^{-1} \left[- \csc \left(\dfrac{\pi}{6}\right)\right] = \csc^{-1} \left[\csc \left(- \dfrac{\pi}{6}\right)\right] = - \dfrac{\pi}{6}$,
where $\;$ $- \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$
$\tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
$\therefore \;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$
$= \dfrac{\pi}{4} + 3 \times \dfrac{\pi}{6} + \dfrac{\pi}{6} - 3 \times \dfrac{\pi}{6}$
$= \dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{5 \pi}{12}$