JEE Main 2021Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestionThe value of tan(2tan−1(35)+sin−1(513))\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)tan(2tan−1(53)+sin−1(135)) is equal to :OptionsA−18169{{ - 181} \over {69}}69−181B22021{{220} \over {21}}21220C−29176{{ - 291} \over {76}}76−291D15163{{151} \over {63}}63151Check AnswerHide SolutionSolution2tan−1(35)=tan−1(6/51−925)=tan−1(651625)=tan−11582{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}2tan−1(53)=tan−1(1−2596/5)=tan−1(251656)=tan−1815 ∴\therefore∴ 2tan−1(35)+sin−1(513)=tan−1(158)+tan−1(512)2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) = {\tan ^{ - 1}}\left( {{{15} \over 8}} \right) + {\tan ^{ - 1}}\left( {{5 \over {12}}} \right)2tan−1(53)+sin−1(135)=tan−1(815)+tan−1(125) =tan−1(158+5121−158,512) = {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)=tan−1(1−815,125815+125) =tan−1(180+4021)=tan−1(22021) = {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)=tan−1(21180+40)=tan−1(21220)