Find the value of ${\tan ^{ - 1}}\left( {\tan \frac{{5\pi }}{6}} \right) + {\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$.

We know that, ${\tan ^{ - 1}}\tan x = x;x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ and ${\cos ^{ - 1}}\cos x = x;x \in [0,\pi ]$

$therefore, {\tan ^{ - 1}}\left( {\tan \frac{{5\pi }}{6}} \right) + {\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$

$= {\tan ^{ - 1}}\left[ {\tan \left( {\pi - \frac{\pi }{6}} \right)} \right] + {\cos ^{ - 1}}\left[ {\cos \left( {\pi + \frac{{7\pi }}{6}} \right)} \right]$

$= {\tan ^{ - 1}}\left( { - \tan \frac{\pi }{6}} \right) + {\cos ^{ - 1}}\left( { - \cos \frac{{7\pi }}{6}} \right)$

$= - {\tan ^{ - 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi - \left[ {{{\cos }^{ - 1}}\cos \left( {\frac{{7\pi }}{6}} \right)} \right]$

and $\left. {{{\cos }^{ - 1}}( - x) = \pi - {{\cos }^{ - 1}}x;x \in [ - 1,1]} \right\}$

$= - {\tan ^{ - 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi - {\cos ^{ - 1}}\left[ {\cos \left( {\pi + \frac{\pi }{6}} \right)} \right]$

$= - {\tan ^{ - 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi - \pi + {\cos ^{ - 1}}\left( {\cos \frac{\pi }{6}} \right)$

$= - \frac{\pi }{6} + 0 + \frac{\pi }{6} = 0$