You seem to have some serious problems with the algebra involved. Part of the problem is failure to use necessary parentheses: the result of applying the quotient rule is
$\frac{2\sin x-(2x+1)\cos x}{\sin^2x}\;,$
where the parentheses around $2x+1$ are absolutely necessary. If you choose to multiply out the numerator, you should get
$\frac{2\sin x-2x\cos x-\cos x}{\sin^2x}\;.$
Alternatively, you can split it into two fractions:
$\begin{align*} \frac{2\sin x-(2x+1)\cos x}{\sin^2x}&=\frac{2\sin x}{\sin^2x}-\frac{(2x+1)\cos x}{\sin^2x}\\\\ &=2\csc x-(2x+1)\cot x\csc x\\\\ &=\csc x\Big(2-(2x+1)\cot x\Big)\;. \end{align*}$