Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$ 
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$

Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x 
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x} \tag1 \\[0.8em]
&=\frac{\cos^2x+2\cos x+1}{\sin^2x} \tag2 \\[0.8em]
&=\frac{1-\sin^2x+1+\dfrac{\sin 2x}{\sin x}}{\sin^2x} \tag3 \\[0.8em]
&=\frac{\;\dfrac{2\sin x-\sin^3x+\sin 2x}{\sin x}\;}{\sin^2x} \tag4 \\[0.8em]
&=\frac{2\sin x-\sin^3x+\sin2 x}{\sin^3x} \tag5
\end{align}$$
I could not prove further to the left hand side from here. I would need help. Thank you in advance.

$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=$$
$$ \frac {2\sin x +2\sin x \cos x }{2\sin x -2\sin x \cos x}=$$
$$\frac {1 + \cos x }{1 -\cos x}=$$
$$\frac {(1 + \cos x)^2 }{1 -\cos^2 x}=$$
$$\frac { 1+2\cos x +\cos ^2 x}{\sin ^2 x}=$$
$$\csc^2x+2\csc x \cot x+\cot^2x$$
