Question
Consider: $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right),\,x \in \left( {0,\frac{\pi }{2}} \right).$$
A normal to $$y = f\left( x \right)$$ at $$x = \frac{p}{6}$$ also passes through the point :
A.
$$\left( {\frac{\pi }{6},0} \right)$$
B.
$$\left( {\frac{\pi }{4},0} \right)$$
C.
(0, 0)
D.
$$\left( {0,\frac{{2\pi }}{3}} \right)$$
Answer :
$$\left( {0,\frac{{2\pi }}{3}} \right)$$
Solution :
$$\eqalign{
& f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right) \cr
& = {\tan ^{ - 1}}\left( {\sqrt {\frac{{{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}}}{{{{\left( {\sin \frac{x}{x} - \cos \frac{x}{2}} \right)}^2}}}} } \right) \cr
& = {\tan ^{ - 1}}\left( {\frac{{1 + \tan \frac{x}{2}}}{{1 - \tan \frac{x}{2}}}} \right) \cr
& = {\tan ^{ - 1}}\left( {\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)} \right) \cr
& \Rightarrow y = \frac{\pi }{4} + \frac{x}{2} \cr
& \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2} \cr
& {\text{Slope of normal }} = \frac{{ - 1}}{{\left( {\frac{{dy}}{{dx}}} \right)}} = - 2 \cr
& {\text{At }}\left( {\frac{\pi }{6},\frac{\pi }{4} + \frac{\pi }{{12}}} \right) \cr
& y - \left( {\frac{\pi }{4} + \frac{\pi }{{12}}} \right) = - 2\left( {x - \frac{\pi }{6}} \right) \cr
& y - \frac{{4\pi }}{{12}} = - 2x + \frac{{2\pi }}{6} \cr
& y - \frac{\pi }{3} = - 2x + \frac{\pi }{3} \cr
& y = - 2x + \frac{{2\pi }}{3} \cr} $$
This equation is satisfied only by the point $$\left( {0,\frac{{2\pi }}{3}} \right)$$