Answer
$$\tan (\pi-\theta) =-\tan (\theta) $$
Work Step by Step
We use the know identities to obtain:
\begin{align*}
\tan (\pi-\theta)&=\frac{\sin (\pi-\theta)}{\cos (\pi-\theta)} \\
&=\frac{\sin (\pi) \cos (\theta)-\cos (\pi) \sin (\theta)}{\cos (\pi) \cos (\theta)+\sin (\pi) \sin (\theta)}\\
&=\frac{0 \times \cos (\theta)-(-1) \times \sin (\theta)}{(-1) \times \cos (\theta)+0 \times \sin (\theta)}\\
&=\frac{0+\sin (\theta)}{-\cos (\theta)+0}=\frac{\sin (\theta)}{-\cos (\theta)}\\
&=-\tan (\theta)
\end{align*}
