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

(N/A) We use the trigonometric identities: $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
$L.H.S. = \frac{\tan (\frac{\pi}{4} + x)}{\tan (\frac{\pi}{4} - x)}$
$= \frac{\frac{\tan \frac{\pi}{4} + \tan x}{1 - \tan \frac{\pi}{4} \tan x}}{\frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x}}$
Since $\tan \frac{\pi}{4} = 1$,we have:
$= \frac{\frac{1 + \tan x}{1 - \tan x}}{\frac{1 - \tan x}{1 + \tan x}}$
$= \frac{1 + \tan x}{1 - \tan x} \times \frac{1 + \tan x}{1 - \tan x} = (\frac{1 + \tan x}{1 - \tan x})^2$
$= R.H.S.$