class 12 maths inverse trigonometric functions

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

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📘 Inverse Trigonometric Functions NCERT Misc. , Q.1 , Page 51 SA

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

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

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

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

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