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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