JEE Main 2016Session, Shift 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002Consider : f (x)=tan−1(1+sinx1−sinx),x∈(0,π2).\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).(x)=tan−1(1−sinx1+sinx),x∈(0,2π). A normal to y=y =y= f(x)\left( x \right)(x) at x=π6x = {\pi \over 6}x=6π also passes through the point:AOption A. (π6,0)\left( {{\pi \over 6},0} \right)(6π,0)BOption B. (π4,0)\left( {{\pi \over 4},0} \right)(4π,0)COption C. (0,0)(0,0)(0,0)DOption D. (0,2π3)\left( {0,{{2\pi } \over 3}} \right)(0,32π)CheckSkip · show solution
