Mark the tick against the correct answer in the following:

To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$

Now, let $x=\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$

$\Rightarrow \cos x=\cos \left(\frac{13 \pi}{6}\right)$

Here, range of principle value of $\cos$ is $[0, \pi]$

$\Rightarrow x=\frac{13 \pi}{6} \notin[0, \pi]$

Hence for all values of $x$ in range $[0, \pi]$, the value of

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

$\Rightarrow \cos x=\cos \left(2 \pi-\frac{\pi}{6}\right)\left(\because \cos \left(\frac{13 \pi}{6}\right)=\cos \left(2 \pi-\frac{\pi}{6}\right)\right)$

$\Rightarrow \cos x=\cos \left(\frac{\pi}{6}\right)(\because \cos (2 \pi-\theta)=\cos \theta)$

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