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. 
 A: Your error lies in how you simplified the right hand side (the denominator specifically).  Try again!  Turn $\tan x$ into $\sin x$ and $\cos x$ with $\displaystyle\tan x =\frac{\sin x}{\cos x}$.
Now,
$$\frac{\tan{x}}{1-\tan^2{x}}=\frac{\left(\frac{\sin x}{\cos x}\right)}{1-\left(\frac{\sin x}{\cos x}\right)^2}$$
You can multiply an expression by 1 and not change the value, (Since $1\cdot a=a$).  Now, the problem is which 1 do you multiply by?
You can achieve this by multiplying and distributing by $\displaystyle \frac{\cos^2x}{\cos^2x}$.
A: $$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$$
$$=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$$
$$=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$$
A: 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
$$
  \frac{ \tan A} {1 - \tan^2 A} .
$$
