Answer
See the verification below.
Work Step by Step
Considered the identity, $\tan \left( \theta +\pi \right)=\tan \theta $
Now, use the sum formulas for tangents $\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ ,
$\tan \left( \theta +\pi \right)=\frac{\tan \theta +\tan \pi }{1-\tan \theta \tan \pi }$
Now put the value $\tan \pi =0$ ,
$\begin{align}
& \tan \left( \theta +\pi \right)=\frac{\tan \theta +\tan \pi }{1-\tan \theta \tan \pi } \\
& =\frac{\tan \theta +0}{1-\tan \theta \cdot 0} \\
& =\frac{\tan \theta }{1} \\
& =\tan \theta
\end{align}$
Hence, the left side is identical to the right side $\tan \left( \theta +\pi \right)=\tan \theta $.
