Answer
$\dfrac {\tan x-1}{\tan x+1}$
Work Step by Step
$\tan \left( \dfrac {3\pi }{4}+x\right) =\dfrac {\tan \dfrac {3\pi }{4}+\tan x}{1-\tan \dfrac {3\pi }{4}\tan x}=\dfrac {-1+\tan x}{1-\left( -1\right) \times \tan x}=\dfrac {\tan x-1}{\tan x+1}$
