Question

Simplify:
${\tan }^{-1}\left[\frac{2\cos x-3\sin x}{3\cos x+2\sin x}\right],where\frac{2}{3}\tan x>-1$

Answer 1

We have, ${\tan }^{-1}\left[\frac{2\cos x-3\sin x}{3\cos x+2\sin x}\right]$

Let $a={\tan }^{-1}\left[\frac{2\cos x-3\sin x}{3\cos x+2\sin x}\right]$

$={\tan }^{-1}\left[\frac{2\cos x(1-\frac{3}{2}\frac{\sin x}{\cos x})}{3\cos x(1+\frac{2}{3}\frac{\sin x}{\cos x})}\right]$

$={\tan }^{-1}\left[\frac{2}{3}\frac{(1-\frac{3}{2}\tan x)}{(1+\frac{2}{3}\tan x)}\right]$

$={\tan }^{-1}[\frac{2}{3}\times \frac{3}{2}\left[\frac{\frac{2}{3}-\tan x}{1+\frac{2}{3}\tan x}\right]]$

$={\tan }^{-1}\left[\frac{\frac{2}{3}-\tan x}{1+\frac{2}{3}\tan x}\right]$

Let $\frac{2}{3}=\tan \alpha ,\alpha ={\tan }^{-1}=\frac{2}{3}$

$={\tan }^{-1}\left[\frac{\tan \alpha -\tan x}{1+\tan \alpha \tan x}\right]$

$={\tan }^{-1}\tan (\alpha -x)$

$=\alpha -x$

$a={\tan }^{-1}\frac{2}{3}-x$

Hence, this is the answer.