Prove that: \[ \frac{\sin 5 A \cos 2 A-\sin 6 A \cos A}{\sin A \sin 2 A-\cos 2 A \cos 3 A}=\tan A \]

Prove that:
\[
\frac{\sin 5 A \cos 2 A-\sin 6 A \cos A}{\sin A \sin 2 A-\cos 2 A \cos 3 A}=\tan A
\]