#### 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

$\tan^{-1}\left ( \tan x \right )=x; x \epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$ and

$\cos^{-1}\left( \cos x \right) = x; x \in \left( 0, \pi \right)$

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

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

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

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

$\left [ since \, \tan^{-1}\left ( -x \right )=- \tan 1 x , x\epsilon R \, and \, \cos^{-1}=\left ( -x \right )=\pi-\cos^{-1}\left ( x \right ), x\epsilon \left ( -1,1 \right ) \right ]$

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

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

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

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

$=0$