Question

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

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

Answer 1

Answer: (D)

f(x)= $tan^{-1} \bigg(\frac{sin x - cos x}{sin x + cos x}\bigg)$
$= tan^{-1} \bigg(\frac{tan x -1 }{tan x +1}\bigg) = tan^{-1} \bigg(tan\bigg(x - \frac{\pi}{4}\bigg)\bigg)$
$\because x - \frac{\pi}{4} \, \in \bigg( - \frac{\pi}{4}, \frac{\pi}{4}\bigg)$
$\therefore \, \, \, f(x) = x - \frac{\pi}{4}$
$\Rightarrow $ its derivative w.r.t. $\frac{x}{2} \, is \, \frac{1}{1/2} = 2$