Question

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

• A. $(0,0)$
• B. $(0,\frac{2\pi }{3})$
• C. $(\frac{\pi }{6},0)$
• D. $(\frac{\pi }{4},0)$

Answer 1

The correct option is B $(0,\frac{2\pi }{3})$

$f\left(x\right)={\tan }^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$ $xϵ(0,\frac{\pi }{2})$ ...Given

Slope at $x=\frac{\pi }{6}$, $y_{\frac{\pi }{6}}={\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$

$f^{\prime }\left(x\right)=\frac{1}{1+\left(\frac{1+\sin x}{1-\sin x}\right)}\left[\frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}}\right]\left[\frac{\cos x(1-\sin x)+\cos x(1+\sin x)}{(1-\sin x{)}^2}\right]$

$f^{\prime }\left(\frac{\pi }{6}\right)=\frac{1}{1+3}\times \frac{1}{2\sqrt{3}}\times \frac{\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{4}}{\frac{1}{4}}$

$=\frac{1}{4}\times \frac{1}{2\sqrt{3}}\times \frac{4\sqrt{3}}{1}=\frac{1}{2}$

$\therefore$ Slope of normal$=-2$

So, equation of line is $\frac{y-\frac{\pi }{3}}{x-\frac{\pi }{6}}=-2$

$(0,\frac{2\pi }{3})$ satisfies this equation $\frac{\frac{-\pi }{3}}{\frac{\pi }{6}}=-2$

So, $(0,\frac{2\pi }{3})$ lies on the line.