JEE Main 2023 · 24 January · Morning Shift · Mathematics · Inverse Trigonometric Functions
C. π3\dfrac{\pi}{3}3π
tan−1(1+33+3)+sec−18+436+33\tan^{- 1}\left(\dfrac{1 + \sqrt{3}}{3 + \sqrt{3}}\right) + \sec^{- 1}\sqrt{\dfrac{8 + 4\sqrt{3}}{6 + 3\sqrt{3}}}tan−1(3+31+3)+sec−16+338+43
=tan−1(1+33(1+3))+sec−14(2+3)3(2+3)= \tan^{- 1}\left(\dfrac{1 + \sqrt{3}}{\sqrt{3}\left(1 + \sqrt{3}\right)}\right) + \sec^{- 1}\sqrt{\dfrac{4\left(2 + \sqrt{3}\right)}{3\left(2 + \sqrt{3}\right)}}=tan−1(3(1+3)1+3)+sec−13(2+3)4(2+3)
=tan−1(13)+sec−143= \tan^{- 1}\left(\dfrac{1}{\sqrt{3}}\right) + \sec^{- 1}\sqrt{\dfrac{4}{3}}=tan−1(31)+sec−134
=tan−1(13)+cos−132=π6+π6=π3= \tan^{- 1}\left(\dfrac{1}{\sqrt{3}}\right) + \cos^{- 1}\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6} + \dfrac{\pi}{6} = \dfrac{\pi}{3}=tan−1(31)+cos−123=6π+6π=3π