Question

Simplify ${\tan }^{-1}\left[\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right]$, if $\frac{a}{b}\tan x>-1$.

Answer 1

${\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right)$

$={\tan }^{-1}\left(\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}\right)$

$={\tan }^{-1}\left(\tan \left({\tan }^{-1}\frac{a}{b}-x\right)\right)$

$={\tan }^{-1}(\frac{a}{b}-x)$