JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 16)
$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :
$${{11\pi } \over {12}}$$
$${{17\pi } \over {12}}$$
$${{31\pi } \over {12}}$$
$$-$$$${{3\pi } \over {4}}$$
Explanation
$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)$$
$$ = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}$$
$$ = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}$$
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