Question

Solve the following,

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

Answer 1

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

LHS:

$=\frac{\frac{\tan \frac{\pi }{4}+tanx}{1-\tan \frac{\pi }{4}\tan x}}{\frac{\tan \frac{\pi }{4}-tanx}{1+\tan \frac{\pi }{4}\tan x}}$

$=\frac{\frac{1+\tan x}{1-1\times \tan x}}{\frac{1-\tan x}{1+1\times \tan x}}$

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

:RHS