Question

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

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

Answer 1

Answer: (D)

f(x)= $ta{n}^{-1}\left(\frac{sinx-cosx}{sinx+cosx}\right)$
$=ta{n}^{-1}\left(\frac{tanx-1}{tanx+1}\right)=ta{n}^{-1}\left(tan\right(x-\frac{\pi }{4}\left)\right)$
$\because x-\frac{\pi }{4}\in (-\frac{\pi }{4},\frac{\pi }{4})$
$\therefore f(x)=x-\frac{\pi }{4}$
$\Rightarrow$ its derivative w.r.t. $\frac{x}{2}is\frac{1}{1/2}=2$