Prove $\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$ So I have this rather simple trigonometric identity that, for the life of me, I cannot solve. I have worked on it for about 2 hours and still can't get it.

Show that 
  $$\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space  \cos A} {\sin A \space+\space \cos A}$$

Here's what I've done so far:
\begin{align}
{2-\csc^2 A \over \csc^2 A+2\cot A} & = {2 - ({1 \over \sin A })^2 \over ({1 \over \sin A })^2 + 2({1 \over \tan A})} \\
&= {{2\sin^2 A \over \sin^2 A} - {1\over \sin^2 A} \over {1\over \sin^2 A} + 2({\cos A \over \sin A})} \\
& = {{2\sin^2 A -1 \over \sin^2 A} \over {1\over sin^2 A}+2({\sin A \cos A \over \sin^2 A})}\\
& = {2 \sin ^2 A -1 \over \sin^2 A} \times {\sin^2 A \over 2\sin A \cos A +1} \\
& = {2\sin^2 A -1 \over 2\sin A \cos A + 1}
\end{align}
Lots of fractions are involved so I fear I may have made a mistake somewhere.

If anyone has any tips on proving these trigonometric identities, could they please add them in their answer? I've been told just to keep trying; though I believe there must be some 'troubleshooting' method to finish the problem.

=$\frac{2 sin^2(A)-1}{1+2sin(A)cos(A)}$
=$\frac{2sin^2(A)-(cos^2(A)+sin^2(A))}{(cos^2(A)+sin^2(A))+2sin(A)cos(A)}$
=$\frac{sin^2(A)-cos^2(A)}{(sin(A)+cos(A))^2}$
=$\frac{(sin(A)+cos(A))(cos(A)+sin(A))}{(sin(A)+cos(A))^2}$
=$\frac{sin(A)-cos(A)}{(sin(A)+cos(A))}$
