Question

Differentiate $\sqrt{\frac{1-\tan x}{1+\tan x}}$

Answer 1

$y=\sqrt{\frac{1+\tan x}{1-\tan x}}$

$y=\sqrt{\frac{1+\tan x}{1-(1\times tanx)}}$

$y=\sqrt{\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}}$

$y=\sqrt{\tan (\frac{\pi }{4}+x)}$

$\frac{dy}{dx}=\frac{1}{2\sqrt{\tan (\frac{\pi }{4}+x)}}\cdot d\left(\tan (\frac{\pi }{4}+x)\right)$

$=\frac{{\sec }^2(\frac{\pi }{4}+x)}{2\sqrt{\tan (\frac{\pi }{4}+x)}}\cdot d(\frac{\pi }{4}+x)$

$\frac{dy}{dx}=\frac{{\sec }^2(\frac{\pi }{4}+x)}{2\sqrt{\tan (\frac{\pi }{4}+x)}}$.