Question

Write the simplest form of an $ta{n}^{-1}\left|\frac{3cosx-4sinx}{4cosx+3sinx}\right|$ if, $\frac{3}{4}tanx>-1$.

Answer 1

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

$[\because \tan (A+B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}]$ $={\tan }^{-1}\left[\tan [\left(\tan \frac{3}{4}\right)-x]\right]$

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