Answer
See proof below.
Work Step by Step
Apply the difference formula:
$\tan(A -B)=\dfrac{\tan A - \tan B}{1 +\tan A \ \tan B }$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS } =\tan(\pi + \theta)
\\=\dfrac{\tan (\pi) -\tan (\theta)}{1+ \tan (\pi) \ \tan (\theta)}
\\=\dfrac{0-\tan (\theta)}{1+ (0) \ \tan (\theta)}
\\=-\tan (\theta)\\=\text{ RHS}$