Find the principal value of each of the following :

$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left(\cos \left(2 \pi+\frac{\pi}{6}\right)\right)$

[ Formula: $\cos (2 \pi+x)=\cos x, \cos$ is positive in the first quadrant. ]

$=\cos ^{-1}\left(\cos \frac{\pi}{6}\right)$ [Formula: $\left.\cos ^{-1}(\cos x)=x\right]$

$=\frac{\pi}{6}$