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.

We have that, using $\sin 2x=2\sin x\cos x$ and cancelling out $2\sin x\neq0$
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}= \dfrac{2\sin x+2\sin x\cos x}{2\sin x-2\sin x\cos x}= \dfrac{1+\cos x}{1-\cos x}$$
then, assuming $\cos x \neq -1$ ( which holds since we need $\sin x\neq0$ for the original expression), multiply by $\frac{1+\cos x} {1+\cos x}$ and simplify to obtain the result using that $1-\cos^2x=\sin^2x$.
