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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 $\cos\Big[\tan^{-1}\frac{a}{b}-x\Big]>0$

$$ \cos(\tan^{-1}\tfrac{a}{b})\cos x+\sin(\tan^{-1}\tfrac{a}{b})\sin x>0\\\implies\cos(\tan^{-1}\tfrac{a}{b})\cos x+\cos(\tan^{-1}\tfrac{a}{b})\tan(\tan^{-1}\tfrac{a}{b})\cos x\tan x>0\\\implies \cos(\tan^{-1}\tfrac{a}{b})\cos x\Big[1+\tan(\tan^{-1}\tfrac{a}{b})\tan x\Big]>0\implies1+\tan(\tan^{-1}\tfrac{a}{b})\tan x>0\\ \implies\tfrac{a}{b}\tan x>-1 $$

Sooraj S
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  • See https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan/1837799#1837799 – lab bhattacharjee Jan 23 '18 at 08:12
  • @labbhattacharjee thanx. but could u pls assist me with my attempt, why am i not able to reach the conditions ? – Sooraj S Jan 23 '18 at 08:42
  • From the logic cited in my link, $$\tan^{-1}\dfrac{\dfrac ab+(-\tan x)}{1-\dfrac ab(-\tan x)}=\tan^{-1}\dfrac ab+\tan^{-1}(-\tan x)$$

    $$\iff\dfrac ab\cdot(-\tan x)<1$$

     – lab bhattacharjee Jan 23 '18 at 08:46
  • @labbhattacharjee pls check i have edited my post. I understand it. thats not my question. How do i reach it proceeding with my attempt ? – Sooraj S Jan 23 '18 at 08:56
  • But $\cos y>0$ does not necessarily imply $$-\dfrac\pi2< y<\dfrac\pi2$$ Another point $$\tan^{-1}(\tan x)$$ not necessarily $$=x$$ right? – lab bhattacharjee Jan 23 '18 at 09:14
  • @labbhattacharjee i think i 4got to mention i am only considering the principal value branch. In that case, $\tan^{-1}\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}=\tan^{-1}\tfrac{a}{b}-x\implies\tfrac{-\pi}{2}<\tan^{-1}\frac{a}{b}-x<\tfrac{\pi}{2}$. And in that range $\cos\big[\tan^{-1}\frac{a}{b}-x\big]>0$, right ? – Sooraj S Jan 23 '18 at 09:21
  • Let us continue this discussion in chat. – lab bhattacharjee Jan 23 '18 at 09:24
  • @labbhattacharjee i am srry. i dont understand how can it be false ? i think reverse of $\cos\big[\tan^{-1}\frac{a}{b}-x\big]>0$ is $\tan^{-1}\frac{a}{b}-x<\pi/2$. But, whats wrong in taking $\tan^{-1}\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}=\tan^{-1}\frac{a}{b}-x\implies -\pi/2<\tan^{-1}\frac{a}{b}-x<\pi/2\implies \cos\big[\tan^{-1}\frac{a}{b}-x\big]>0$ and proceeding further with the poof ? – Sooraj S Jan 23 '18 at 10:23

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