Find the values of

We know that $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi]$, which is the principal value branch of $\cos ^{-1} x$.

Here, $\frac{7 \pi}{6} \notin x \in[0, \pi]$.

Now, $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ can be written as:

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

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

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[-\cos \left(\pi-\frac{5 \pi}{6}\right)\right]$    

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[-\left\{-\cos \left(\frac{5 \pi}{6}\right)\right\}\right]$   [as. $\cos (\pi-\theta)=-\cos \theta]$

$\therefore \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right)=\frac{5 \pi}{6}$

The correct answer is B.