Consider $f(x)=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), x \in\left(0, \frac{\pi}{2}\right)$. A normal to $y=f(x)$ at $x=\frac{\pi}{6}$ also passes through the point:

(A) $\left(0, \frac{2 \pi}{3}\right)$

(B) $\left(\frac{\pi}{6}, 0\right)$

(C) $\left(\frac{\pi}{4}, 0\right)$

JEE Main · 2016 · 03 Apr 2016 · Offline · Application of Derivatives

Consider

f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})

. A normal to

y = f (x)

x = \frac{π}{6}

also passes through the point:

Answer: (A)

(0, \frac{π}{2})

the argument simplifies:

\frac{1 + s i n x}{1 - s i n x} = tan (\frac{π}{4} + \frac{x}{2})

, so

f (x) = \frac{π}{4} + \frac{x}{2}

and

f^{'} (x) = \frac{1}{2}

. At

x = \frac{π}{6}

f = \frac{π}{4} + \frac{π}{12} = \frac{π}{3}

, and the normal has slope

- \frac{1}{f ^{'}} = - 2

y - \frac{π}{3} = - 2 (x - \frac{π}{6})

. Setting

x = 0

gives

y = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3}

, so the normal passes through

(0, \frac{2 π}{3})

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option