Answer
See the explanation below.
Work Step by Step
By using identities,
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{sinx}{\cos x}$.
Left side can be written as:
$\begin{align}
& \frac{2(\tan x-\cot x)}{{{\tan }^{2}}x-{{\cot }^{2}}x}=\frac{2(\tan x-\cot x)}{\left( \tan x+\cot x \right)\left( \tan x-\cot x \right)} \\
& =\frac{2}{\left( \tan x+\cot x \right)} \\
& =\frac{2}{\left( \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \right)}\times \frac{\sin x\cos x}{\sin x\cos x} \\
& =\frac{2\sin x\cos x}{{{\cos }^{2}}x+{{\sin }^{2}}x}
\end{align}$
By using the trigonometric identities: ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $2\sin x\cos x=\sin 2x$
Simplifying it further,
$\begin{align}
& \frac{2\sin x\cos x}{{{\cos }^{2}}x+{{\sin }^{2}}x}=\frac{\sin 2x}{1} \\
& =\sin 2x
\end{align}$
Thus, the left side of the equation is equal to $\sin 2x$.
Hence, the left side is equal to $\sin 2x$.
