Question

Write the following functions in the simplest form :
${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),0<x<\pi$

Answer 1

Given : ${\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
Dividing by $\cos x$ in numerator and denominator.
$\Rightarrow {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{\cos x}}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\cdot \tan x}\right)$
Using $\tan (A-B)$ formula, we get
$\Rightarrow {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)={\tan }^{-1}\left(\tan (\frac{\pi }{4}-x)\right)$
$\therefore {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi }{4}-x$