Answer
$$\tan(\pi-x)=-\tan x$$
Work Step by Step
$$\tan(\pi-x)$$
Apply the identity of tangent of a difference here, we have $$=\frac{\tan\pi-\tan x}{1+\tan\pi\tan x}$$
We know that $\tan\pi=0$. If you do not already know, see that $\tan\pi=\frac{\sin\pi}{\cos\pi}$, in which $\sin\pi$ equals $0$ and $\cos\pi$ equals $1$. That means,
$$\tan(\pi-x)=\frac{0-\tan x}{1-0\times \tan x}$$ $$=\frac{-\tan x}{1}$$ $$=-\tan x$$
