Question
Prove that:
$\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$
$\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$