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