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JEE Main 2021
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

The value of tan(2tan1(35)+sin1(513))\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right) is equal to :

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Solution

2tan1(35)=tan1(6/51925)=tan1(651625)=tan11582{\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} \therefore 2tan1(35)+sin1(513)=tan1(158)+tan1(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) =tan1(158+5121158,512) = {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right) =tan1(180+4021)=tan1(22021) = {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)

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