Inverse Trigonometric Functions — Class 12 Maths Solution

We have, ${\cos ^{ - 1}}\left( {\cos \frac{{14\pi }}{3}} \right) = {\cos ^{ - 1}}\cos \left( {4\pi + \frac{{2\pi }}{3}} \right)$

$= {\cos ^{ - 1}}\cos \frac{{2\pi }}{3}$
$= \frac{{2\pi}}{3}$

Note
Remember that, ${\cos ^{ - 1}}\left( {\cos \frac{{14\pi }}{3}} \right) \ne \frac{{14\pi }}{3}$
Since $,\frac{{14\pi }}{3} \notin [0,\pi ]$