Question

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

Answer 1

$L.H.S=\frac{\tan (\frac{\pi }{4}+x)}{\tan (\frac{\pi }{4}-x)}$
$\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\cdot \tan B}$
$=\frac{\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}}{\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}}=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}$
$=\frac{1+\tan x}{1-\tan x}\times \frac{1+\tan x}{1-\tan x}={\left(\frac{1+\tan x}{1-\tan x}\right)}^2$
L.H.S. = R.H.S.
Hence proved.