QJEEM21Ma6c52: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6c52 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)

is equal to :

Accepted Answer

2{\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

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)

= {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)

= {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)

Step-by-Step Verified Solution