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?
 A: Try working with the left side instead. We have
$$\csc^2 x-\csc x = \csc x(\csc x - 1) = \csc x\left(\frac{1}{\sin x}-1\right).$$
Making a common denominator, we have
$$\csc^2 x-\csc x = \csc x\frac{1-\sin x}{\sin x}.$$
We want $\frac{1}{1+\sin x}$ somehow (based on the right side), so maybe multiplying and dividing by $1+\sin x$ might do the trick.. (You can do this since multiplying by $\frac{1+\sin x}{1+\sin x}$ does not change anything..)
A: $$\frac{1}{sin^2x}-\frac{1}{sinx}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}-\frac{sinx}{sin^2x}=\frac{\frac{cos^2x}{sin^2x}}{1+sinx}$$
$$\frac{1}{sin^2x}(1-{sinx})=\frac{1}{sin^2x}(\frac{cos^2x}{1+sinx})$$
$$\frac{1}{sin^2x}(1-{sinx})(1+sinx)=\frac{1}{sin^2x}{cos^2x}$$
$$\frac{1}{sin^2x}(1-{sin^2x})=\frac{1}{sin^2x}{cos^2x}$$ 
$$\frac{1}{sin^2x}(cos^2x)=\frac{1}{sin^2x}{cos^2x}$$ 
And we're done.
A: We have $$\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)$$
$$\implies \csc^2x-\csc x=\frac{1-\sin x}{\sin^2x}=\frac{\cos^2x}{\sin^2x(1+\sin x)}=\cdots$$
