To answer this question we need to remember the identity:
[tex]\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}[/tex]and that
[tex]\tan\frac{3\pi}{4}=-1[/tex]If we let A be 3π/4 and B be x, we have:
[tex]\begin{gathered} \tan(\frac{3\pi}{4}+x)=\frac{\tan\frac{3\pi}{4}+\tan x}{1-\tan\frac{3\pi}{4}\tan x} \\ =\frac{-1+\tan x}{1-(-1)\tan x} \\ =\frac{-1+\tan x}{1+\tan x} \end{gathered}[/tex]Therefore, we have that:
[tex]\frac{-1+\tan x}{1+\tan x}=\tan(\frac{3\pi}{4}+x)[/tex]