Question

Write the simplest form of ${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),0<x<\frac{\pi }{2}$.

Answer 1

Given:
${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\right)$
$={\tan }^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\cdot \tan x}\right)$
$={\tan }^{-1}\left(\tan (\frac{\pi }{4}-x)\right)$
$\therefore {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi }{4}-x$.