Question

The derivative of ${\tan }^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right),$ with respect to $\frac{x}{2},$ where $(x\in (0,\frac{\pi }{2}))$ is:

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $2$

$y={\tan }^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)\Rightarrow y={\tan }^{-1}\left(\frac{\tan x-1}{\tan x+1}\right)\Rightarrow y=-{\tan }^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\Rightarrow y=-{\tan }^{-1}\left[\tan (\frac{\pi }{4}-x)\right]\because 0<x<\frac{\pi }{2}\Rightarrow -\frac{\pi }{2}<-x<0\Rightarrow -\frac{\pi }{4}<\frac{\pi }{4}-x<\frac{\pi }{4}\Rightarrow y=-(\frac{\pi }{4}-x)\Rightarrow y=-\frac{\pi }{4}+x\frac{dy}{d\left(x/2\right)}=\frac{1}{\left(1/2\right)}=2$