JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestion2π\pi π - (sin−145+sin−1513+sin−11665)\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)(sin−154+sin−1135+sin−16516) is equal to :OptionsA7π4{{7\pi } \over 4}47πB5π4{{5\pi } \over 4}45πC3π2{{3\pi } \over 2}23πDπ2{\pi \over 2}2πCheck AnswerHide SolutionSolution2π−(sin−145+sin−1513+sin−11665)2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)2π−(sin−154+sin−1135+sin−16516) =2π−(tan−143+tan−1512+tan−11663) = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)=2π−(tan−134+tan−1125+tan−16316) =2π−{tan−1(43+5121−43.512)+tan−11663} = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}=2π−{tan−1(1−34.12534+125)+tan−16316} =2π−(tan−16316+tan−11663) = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)=2π−(tan−11663+tan−16316) = 2π−(tan−16316+cot−16316)2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)2π−(tan−11663+cot−11663) =2π−π2 = 2\pi - {\pi \over 2}=2π−2π =3π2 = {{3\pi } \over 2}=23π