Question:
Prove that
$\tan \left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}$
Solution:
In this question the following formulas will be used:
$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
$\tan \left(\frac{\pi}{4}-x\right)=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}$
$=\frac{1-\tan x}{1+\tan x} \because \tan \frac{\pi}{4}=1$