JEE Main 2021Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionThe value of tan−1(cos(15π4)−1sin(π4)){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)tan−1(sin(4π)cos(415π)−1) is equal to :OptionsA−π4 - {\pi \over 4}−4πB−π8 - {\pi \over 8}−8πC−5π12 - {{5\pi } \over {12}}−125πD−4π9 - {{4\pi } \over 9}−94πCheck AnswerHide SolutionSolutiontan−1(cos(15π4)−1sinπ4){\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)tan−1(sin4πcos(415π)−1) =tan−1(12−112) = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)=tan−1(2121−1) =tan−1(1−2)=−tan−1(2−1) = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)=tan−1(1−2)=−tan−1(2−1) =−π8 = - {\pi \over 8}=−8π