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JEE Main 2019
Inverse Trigonometric Functions
Inverse Trigonometric Functions
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Question

sin1(sin2π3)+cos1(cos7π6)+tan1(tan3π4){\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 :

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Solution

sin1(32)+cos1(32)+tan1(1){\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right) =π3+5π6π4 = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4} =4π+10π3π12=11π12 = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}

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