A) \[\left( \frac{\pi }{4},0 \right)\]
B) \[\left( 0,\text{ }0 \right)\]
C) \[\left( 0,\frac{2\pi }{3} \right)\]
D) \[\left( \frac{\pi }{6},0 \right)\]
Correct Answer: C
Solution :
\[f(x)={{\tan }^{-1}}\left( \sqrt{\frac{1+\sin x}{1-\sin x}} \right)\]where\[x\in \left( 0,\frac{\pi }{2} \right)\] \[={{\tan }^{-1}}\left( \sqrt{\frac{{{(1+\sin x)}^{2}}}{1-{{\sin }^{2}}x}} \right)\] \[={{\tan }^{-1}}\left( \frac{1+\sin {{x}^{2}}}{|\cos x|} \right)\] \[={{\tan }^{-1}}\left( \frac{1+\sin {{x}^{2}}}{\cos x} \right)\]\[\left( as\,x\in \left( 0,\frac{\pi }{2} \right) \right)\] \[={{\tan }^{-1}}\left( \frac{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}} \right)\] \[={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}+\frac{x}{2} \right) \right)\] \[f(x)=\frac{\pi }{4}+\frac{x}{2}\]as\[x\in \left( 0,\frac{\pi }{2} \right)\Rightarrow f'\left( \frac{\pi }{6} \right)=\frac{1}{2}\] \[\therefore \]Equation of normal \[\left( y-\frac{\pi }{3} \right)=-2\left( x-\frac{\pi }{6} \right)\]which passes through\[\left( 0,\frac{2\pi }{3} \right)\]
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