class 12 maths inverse trigonometric functions

${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]$

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📘 Inverse Trigonometric Functions NCERT Ex. 2.2, Q. 11 , Page 48 SA

${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]$

Official Solution

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We have
${\tan ^{ - 1}}\left\{ {2\cos \left( {2{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right)} \right\} = {\tan ^{ - 1}}\left\{ {2\cos \left( {2 \times \frac{\pi }{6}} \right)} \right\}$

$= {\tan ^{ - 1}}\left\{ {2\cos \frac{\pi }{3}} \right\} = {\tan ^{ - 1}}\left[ {2 \times \frac{1}{2}} \right] = {\tan ^{ - 1}}1 = \frac{\pi }{4}$

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