Prove that

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:

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

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

$=\frac{1-\tan x}{1+\tan x} \because \tan \frac{\pi}{4}=1$

 

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