$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :
Solution
<p>$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)$$</p>
<p>$= {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}$</p>
<p>$= {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Properties and Identities
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