${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)$ is equal to

(A) $\frac{{7\pi }}{6}$

(B) $\frac{{5\pi }}{6}$

(C) $\frac{\pi }{3}$

(D) $\frac{\pi }{6}$

Option B is correct

${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) \ne \frac{{7\pi }}{6}$ as principal value branch of ${\cos ^{ - 1}}\;\,is\;\;[0,\;\pi ]$

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

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

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