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JEE Main 2019
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

tan(2tan115+sec152+2tan118)\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right) is equal to :

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Solution

tan(2tan115+sec152+2tan118)\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right) =tan(2tan1(15+18115.18)+sec152) = \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[2tan113+tan112] = \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right] =tan[tan123119+tan112] = \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right] =tan[tan134+tan112] = \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right] =tan[tan134+12138]=tan[tan15458] = \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[tan12]=2 = \tan [{\tan ^{ - 1}}2] = 2

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