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Prove that $\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2$
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$\begin{aligned} \text { L.H.S. } &=\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}} \\ &=\left(\frac{1+\tan x}{1-\tan x}\right)^2=\text { R.H.S. } \end{aligned}$
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