Question

Write the simplest form of $ta{n}^{-1}\left(\frac{cosx-sinx}{cosx+sinx}\right),0<x<\frac{\pi }{2}$.

Answer 1

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

$={\tan }^{-1}\left(\frac{(1-\tan x)}{(1+\tan x)}\right)x\in (0,\frac{\pi }{2})\Rightarrow 1=\tan \frac{\pi }{4}$

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

$\therefore$ ${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi }{4}-x$ if $0<x<\frac{\pi }{2}$