Question

Consider $f\left(x\right)$ $=ta{n}^{-1}$$\left(\sqrt{\frac{1+sinx}{1-sinx}}\right)$,$xϵ(0,\frac{\pi }{2})$. A normal to $y=f\left(x\right)atx=\frac{\pi }{6}$ also passes through the point :

• A.
• B.
• C.
• D. ( 0, 0 )

Answer 1

The correct option is A

$f\left(x\right)=ta{n}^{-1}\sqrt{\frac{1+sinx}{1-sinx}}$

$=ta{n}^{-1}\sqrt{\frac{{(cos\frac{x}{2}+sin\frac{x}{2})}^2}{{(cos\frac{x}{2}-sin\frac{x}{2})}^2}}$

$=ta{n}^{-1}\left(tan(\frac{\pi }{4}+\frac{x}{2})\right)$

$=\frac{\pi }{4}+\frac{x}{2}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}andatx=\frac{\pi }{6},f\left(x\right)=\frac{\pi }{3}$

So, equation of normal is

$y-\frac{\pi }{3}=-2(x-\frac{\pi }{6})\Rightarrow y+2x=\frac{2\pi }{3}$