Discussion Forum

 Consider $f(x)= tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},x\epsilon(0,\frac{\pi}{2})$. A normal to y= f(x) at x= $\frac{\pi}{6}$ also passes through the point

Answer:

Explanation:

We have

$f(x)= tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},x\epsilon(0,\frac{\pi}{2})$

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

   $f(x)= tan^{-1}({\frac{\cos\frac{x}{2}+\sin\frac{x}{2}^{}}{\cos\frac{x}{2}-\sin \frac{x}{2}^{}}})$

(  $\cos \frac{x}{2}>\sin \frac{x}{2} for$  $0< \frac{x}{2}. \frac{\pi}{4}$ 

$=\tan^{-1}(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})$

$=\tan^{-1}[\tan(\frac{\pi}{4}+\frac{x}{2})]=\frac{\pi}{4}+\frac{x}{2}$

$\Rightarrow f'(x)=\frac{1}{2}\Rightarrow f'(\frac{\pi}{6})=\frac{1}{2}$

  Now, equation of normal at x= $\frac{\pi}{6}$ is given by

  $(y-f(\frac{\pi}{6}))=-2(x-\frac{\pi}{6})$

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

$[\because f(\frac{\pi}{6})=\frac{\pi}{4}+\frac{\pi}{12}=\frac{4\pi}{12}=\frac{\pi}{3}$

which passes  through (0, $\frac{2\pi}{3}$)