Answer
See the explanation below.
Work Step by Step
To verify the given identity,
$\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$
Recall the Trigonometric Identities and apply below,
$\begin{align}
& \sin 2\theta =2\sin \theta \cos \theta \\
& \tan \theta =\frac{\sin \theta }{\cos \theta } \\
& \cos \theta =\frac{1}{\text{sec}\theta } \\
& 1+{{\tan }^{2}}\theta =\text{se}{{\text{c}}^{2}}\theta \\
\end{align}$
Consider the left side of the given expression,
$\sin 2\theta =2\sin \theta \cos \theta $
Multiply and divide by $\cos \theta $.
$\begin{align}
& \sin 2\theta =2\sin \theta \cos \theta \times \frac{\cos \theta }{\cos \theta } \\
& =2\frac{\sin \theta }{\cos \theta }\times {{\cos }^{2}}\theta \\
& =2\tan \theta {{\cos }^{2}}\theta \\
& =2\tan \theta \frac{1}{{{\sec }^{2}}\theta }
\end{align}$
Substitute $1+{{\tan }^{2}}\theta $ for $\text{se}{{\text{c}}^{2}}\theta $.
$\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$
Hence, it is proved that the given identity $\sin 2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }$ holds true.
