JEE Main 2019Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionsin1(sin2π3)+cos−1(cos7π6)+tan−1(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)sin1(sin32π)+cos−1(cos67π)+tan−1(tan43π) is equal to :OptionsA11π12{{11\pi } \over {12}}1211πB17π12{{17\pi } \over {12}}1217πC31π12{{31\pi } \over {12}}1231πD-$$$${{3\pi } \over {4}}Check AnswerHide SolutionSolutionsin−1(32)+cos−1(−32)+tan−1(−1){\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)sin−1(23)+cos−1(2−3)+tan−1(−1) =π3+5π6−π4 = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}=3π+65π−4π =4π+10π−3π12=11π12 = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}=124π+10π−3π=1211π