Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$\begin{aligned} \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] & = \sin^{-1} \left[\sin \left(\pi - \dfrac{\pi}{6}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{6} \notin \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \\\\ & = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] & \\\\ & = \dfrac{\pi}{6} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \end{aligned}$

$\begin{aligned} \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right)\right] & = \cos^{-1} \left[\cos \left(2 \pi - \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{3} \notin \left[0, \pi\right] \right\} \\\\ & = \cos^{-1} \left[\cos \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left[0, \pi\right] \right\} \end{aligned}$

$\begin{aligned} \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right] & = \tan^{-1} \left[\tan \left(2 \pi + \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{7 \pi}{3} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \\\\ & = \tan^{-1} \left[\tan \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \end{aligned}$

$\therefore \; \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$= \dfrac{\pi}{6} + \dfrac{\pi}{3} + \dfrac{\pi}{3}$

$= \dfrac{5 \pi}{6}$