Question

Prove that:

\[\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A\]

Answer 1

Consider LHS: 
\[ \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}\]
\[ = \frac{\sin 3A + \sin 9A + \sin 5A + \sin 7A}{\cos 3A + \cos 9A + \cos 5A + \sin 7A}\]
\[ = \frac{2\sin \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\sin \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}{2\cos \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\cos \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}\]
\[ = \frac{2\sin 6A \cos \left( - 3A \right) + 2\sin 6A \cos \left( - A \right)}{2\cos 6A \cos \left( - 3A \right) + 2\cos 6A \cos \left( - A \right)}\]
\[ = \frac{2\sin 6A \cos 3A + 2\sin 6A \cos A}{2\cos 6A \cos 3A + 2\cos 6A \cos A}\]
\[ = \frac{2\sin 6A\left[ \cos 3A + \cos A \right]}{2\cos 6A\left[ \cos 3A + \cos A \right]}\]
\[ = \tan 6A\]
 = RHS
Hence, LHS = RHS.