Question

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

Answer 1

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

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

$={\tan }^{-1}\left[\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}\right]$

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

$=\frac{\pi }{4}+x$