Question

If $y={\tan }^{-1}\left[\frac{a\cos x+b\sin x}{b\cos x-a\sin x}\right]$, find$\frac{dy}{dx}$.

Answer 1

$\begin{array}{c}{\tan }^{-1}\left[\frac{a\cos x+b\sin x}{b\cos x-a\sin x}\right]\\ {\tan }^{-1}\left[\frac{\frac{a\cos x}{b\cos x}+\frac{b\sin x}{bcosx}}{\frac{b\cos x}{b\cos x}-\frac{a\sin x}{b\cos x}}\right]\\ {\tan }^{-1}\left[\frac{\frac{a}{b}+\tan x}{1-\frac{a}{b}\tan x}\right]\\ {\tan }^{-1}\left(\frac{a}{b}\right)+{\tan }^{-1}\left(\tan x\right)\\ {\tan }^{-1}\left(\frac{a}{b}\right)+x\end{array}$