Question

Simplest form of: ${\tan }^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)$

Answer 1

${\tan }^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)$

${\tan }^{-1}\left[\frac{4\cos x(\frac{3}{4}-\frac{\sin x}{\cos x})}{4\cos x(1+\frac{3}{4}\frac{\sin x}{\cos x})}\right]$

Let $\tan \theta =\frac{3}{4}$

$\Rightarrow {\tan }^{-1}\left[\frac{\tan \theta -\tan x}{1+\tan \theta \cdot \tan x}\right]$

$\Rightarrow {\tan }^{-1}\left(\tan \right(\theta -x\left)\right)$

$\Rightarrow \theta -x={\tan }^{-1}\frac{3}{4}-x$