Question

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

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

Answer 1

Answer: (B)

$f\left(x\right) = \tan^{-1} \tan \left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{\pi}{4} + \frac{x}{2}$
If $ x = \frac{\pi}{6} , f \frac{\pi}{6} = \frac{\pi}{4} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$
$ f'\left(x\right) = \frac{1}{2} $
Normal $y - \frac{\pi}{3} =- 2 \left( x - \frac{\pi}{6}\right) $
So, $ \left(0, \frac{2\pi}{3}\right) $ Satisfy