We must show that $\frac{\sin 5A}{\sin A} - \frac{\cos 5A}{\cos A} = 4 \cos 2A$. We use the trigonometric identities to show that the Left side of the equation is equal to the Right side of the equation.

LHS

$=\frac{\sin 5A}{\sin A} - \frac{\cos 5A}{\cos A}$

Take the LCM and simplify,

$=\frac{\sin 5A \cdot \cos A - \cos 5A \cdot \sin A}{\sin A \cdot \cos A}$

Use the trigonometric identity for $\sin (A-B) = \sin A \cos B - \cos A \sin B$.

$=\frac{\sin (5A -A)}{\sin A \cdot \cos A}$

$=\frac{\sin (4A)}{\sin A \cdot \cos A}$

$=\frac{\sin (2*2A)}{\sin A \cdot \cos A}$

$=\frac{2 \sin 2A \cdot \cos 2A}{\sin A \cdot \cos A}$

Multiply and divide the expression by 2, we get,

$=\frac{2 \cdot 2 \sin 2A \cdot \cos 2A}{2sinAcosA}$

$=\frac{4 \sin 2A \cdot \cos 2A}{\sin 2A}$

= 4 cos2A

RHS

Hence, we have shown that $\frac{\sin 5A}{\sin A} - \frac{\cos 5A}{\cos A} = 4 \cos 2A$.