Simplify
$$ \tan^{-1}\frac{a\cos x-b\sin x}{b\cos x+a\sin x},\quad\text{ if }\quad\frac{a}{b}\tan x>-1 $$
I understand the solution: $$ L.H.S=\tan^{-1}\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}=\tan^{-1}\big[\tan(\tan^{-1}\tfrac{a}{b}-x)\big]=\tan^{-1}\frac{a}{b}-x $$ Alright, but what about the conditions $\frac{a}{b}\tan x>-1$ ?
My Attempt:
$$ \frac{-\pi}{2}<\tan^{-1}\frac{a}{b}-x<\frac{\pi}{2}\implies \frac{-\pi}{2}<x-\tan^{-1}\frac{a}{b}<\frac{\pi}{2}\\\implies \tan^{-1}\frac{a}{b}-\frac{\pi}{2}<x<\tan^{-1}\frac{a}{b}+\frac{\pi}{2}\\\implies \tan\Big[\tan^{-1}\frac{a}{b}-\frac{\pi}{2}\Big]<\tan x<\tan\Big[\tan^{-1}\frac{a}{b}+\frac{\pi}{2}\Big]\\\implies -\tan\big[\tan^{-1}\frac{a}{b}\big] <\tan x<-\tan\big[\tan^{-1}\frac{a}{b}\big]$$
How do I proceed further and reach the given condition ?
Note 1: We are only considering the pricipal value branch
Note 2: I don't want to derive the conditions from the fact that $\cos\Big[\tan^{-1}\frac{a}{b}-x\Big]>0$
$$ \cos(\tan^{-1}\tfrac{a}{b})\cos x+\sin(\tan^{-1}\tfrac{a}{b})\sin x>0\\\implies\cos(\tan^{-1}\tfrac{a}{b})\cos x+\cos(\tan^{-1}\tfrac{a}{b})\tan(\tan^{-1}\tfrac{a}{b})\cos x\tan x>0\\\implies \cos(\tan^{-1}\tfrac{a}{b})\cos x\Big[1+\tan(\tan^{-1}\tfrac{a}{b})\tan x\Big]>0\implies1+\tan(\tan^{-1}\tfrac{a}{b})\tan x>0\\ \implies\tfrac{a}{b}\tan x>-1 $$