Question

Differentiate w.r.t. $x$
${\tan }^{-1}\left(\frac{a\cos x-b\sin x}{a\cos x+b\sin x}\right)$

Answer 1

$\begin{array}{c}y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{a\cos x+b\sin x}\right)\\ ={\tan }^{-1}\left(\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\times \tan x}\right)\\ Put,\frac{a}{b}=\tan \alpha \\ y={\tan }^{-1}\left(\frac{\tan \alpha -\tan x}{1+\tan \alpha \tan x}\right)\\ ={\tan }^{-1}\left(\tan \left(\alpha -x\right)\right)=\alpha -x\\ y={\tan }^{-1}\left(\frac{a}{b}\right)-x\\ \frac{dy}{dx}=0-1=-1\end{array}$