How to solve this trigonometric identity? I have tried it several times and I do not know what I am doing wrong. This is the identity:
$$\csc^2x - \csc x = \frac{\cot^2 x}{1 + \sin x}$$
Trying with RHS, I get: 
$$\frac{\cos^2 x }{ \sin^2 x} + \frac{\cos^2 x }{ \sin x}$$
With LHS: 
$$\frac{1 }{ \sin^2 x} - \frac{\sin x }{ \sin^2 x}$$
But I do not know how to continue after that. Am I missing something?

$$\frac{1}{sin^2x}-\frac{1}{sinx}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}-\frac{sinx}{sin^2x}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}(1-{sinx})=\frac{1}{sin^2x}(\frac{cos^2x}{1+sinx})$$
$$\frac{1}{sin^2x}(1-{sinx})(1+sinx)=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(1-{sin^2x})=\frac{1}{sin^2x}{cos^2x}$$ 
$$\frac{1}{sin^2x}(cos^2x)=\frac{1}{sin^2x}{cos^2x}$$ 
And we're done.
