Inverse Trigonometric Functions — Class 12 Maths Solution

${\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right) \ne \frac{{13\pi }}{6}$ as the range of principal value branch of ${\cos ^{ - 1}}$ is $[0,\;\pi ]$

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

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