JEE Main 2019Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestiontan(2tan−115+sec−152+2tan−118)\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)tan(2tan−151+sec−125+2tan−181) is equal to :OptionsA1B2C14\frac{1}{4}41D54\frac{5}{4}45Check AnswerHide SolutionSolutiontan(2tan−115+sec−152+2tan−118)\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)tan(2tan−151+sec−125+2tan−181) =tan(2tan−1(15+181−15 . 18)+sec−152) = \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(2tan−1(1−51.8151+81)+sec−125) =tan[2tan−113+tan−112] = \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right]=tan[2tan−131+tan−121] =tan[tan−1231−19+tan−112] = \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right]=tan[tan−11−9132+tan−121] =tan[tan−134+tan−112] = \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right]=tan[tan−143+tan−121] =tan[tan−134+121−38]=tan[tan−15458] = \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−11−8343+21]=tan[tan−18545] =tan[tan−12]=2 = \tan [{\tan ^{ - 1}}2] = 2=tan[tan−12]=2