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