Question

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

Answer 1

Answer: 2

$f\left(x\right)=ta{n}^{-1}\left(\frac{tanx-1}{tanx+1}\right)$
$=-ta{n}^{-1}\left(tan\left(\frac{\pi }{4}-x\right)\right)\left[\because \frac{\pi }{4}-x\in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)\right]$
So, $f\left(x\right)=-\left(\frac{\pi }{4}-x\right)=x-\frac{\pi }{4}$
Let $y=\frac{x}{2}$
$\Rightarrow f\left(y\right)=2y-\frac{\pi }{4}$
Now, differentiate w.r.t. $y,\frac{df(y)}{dy}=2$.