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.
 A: 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 $1-\cos^2x=\sin^2x$.
A: The following relations would be used in the answer:$$1-\cos^2x = \sin^2x$$ $$\sin 2x=2\sin x\cos x$$
Taking LHS,
$$\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x} = \frac{2\sin x(1+\cos x)}{2\sin x(1-\cos x)}$$
Cancelling $\sin 2x$
$$=\frac{1+\cos x}{1-\cos x}$$
Multiplying $1+\cos x$ to numerator and denominator
$$=\frac{(1+\cos x)^2}{1-\cos^2x}$$
which is equivalent to $$=\frac{(1+\cos^2x+2\cos x)}{\sin^2x}$$
Divinding each term by denominator, we get LHS as $$=\csc^2x+2\csc x\cot x+\cot^2x$$
which is the RHS.
Hence, Proved.
A: $$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=$$
$$ \frac {2\sin x +2\sin x \cos x }{2\sin x -2\sin x \cos x}=$$
$$\frac {1 + \cos x }{1 -\cos x}=$$
$$\frac {(1 + \cos x)^2 }{1 -\cos^2 x}=$$
$$\frac { 1+2\cos x +\cos ^2 x}{\sin ^2 x}=$$
$$\csc^2x+2\csc x \cot x+\cot^2x$$
