Prove that

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Question:

Prove that $\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}$

 

Solution:

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

$\Rightarrow \frac{\frac{1+\tan x}{1-1 \cdot \tan x}}{\frac{1-\tan x}{1+1 \cdot \tan x}}=\frac{1+\tan x}{1-\tan x} \cdot \frac{1+\tan x}{1-\tan x}$

$\Rightarrow\left(\frac{1+\tan x}{1-\tan x}\right)^{2}$

Hence, Proved.

 

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