Prove $\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}$ How would I simplify this difficult trigonometric identity:
$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$ 
I am not exactly sure what to do.
I simplified the right side to
$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$
But how would I proceed. 

Use $\sin 2A= 2\sin A \cos A$ and 
$ \cos 2A= \cos^2 A- \sin^2 A$ to get
$$
\frac{\sin A \cos A}{\cos^2 A- \sin^2 A}=\frac{\sin 2A}{2\cos 2A}=\frac12\tan{2A},
$$
which is equivalent to
$$
  \frac{ \tan A} {1 - \tan^2 A} .
$$
