QJEEM22Mac706: JEE Main 2022 Mathematics - Class 12 JEEM22Mac706 Solution

$\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)$

$= \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)$

$= \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$= \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$= \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$= \tan \left[ {{{\tan }^{ - 1}}{{{3 \over 4} + {1 \over 2}} \over {1 - {3 \over 8}}}} \right] = \tan \left[ {{{\tan }^{ - 1}}{{{5 \over 4}} \over {{5 \over 8}}}} \right]$

$= \tan [{\tan ^{ - 1}}2] = 2$