MathematicsMedium66×since 2002Q4079tan−1(1+33+3)+sec−1(8+436+33){\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)tan−1(3+31+3)+sec−16+338+43 is equal to :Aπ2{\pi \over 2}2πBπ3{\pi \over 3}3πCπ6{\pi \over 6}6πDπ4{\pi \over 4}4πCheck answerSkip