Answer
True.
Work Step by Step
Recall:
$\tan(\alpha+\beta)=\dfrac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$
Hence,
$\tan(2\theta)\\
=\tan(\theta+\theta)\\
=\dfrac{\tan(\theta)+\tan(\theta)}{1-\tan(\theta)\tan(\theta)}\\
=\dfrac{2\tan(\theta))}{1-\tan^2(\theta)}$
Thus, the given statement is true.