Answer:
See Explanation
Step-by-step explanation:
[tex] \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ LHS = \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} \\ \\ = \frac{ \sin5A \:\cos A - \cos5A \: \sin A}{\sin A \:\cos A } \\ \\ = \frac{ \sin(5A -A )}{\sin A \:\cos A} \\ \\ = \frac{ \sin 4A}{\sin A \:\cos A} \\ \\ = \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ =4\cos 2A \\ \\ = RHS \\ \\ thus \\ \\ \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ hence \: proved[/tex]
