Question

Prove that:

\[\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A} = \tan 8A\]

Answer 1

Consider LHS: 
\[ \frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 2A \sin A + \cos 6A \sin 3A}\]
Multiplying numerator and denominator by 2, we get
\[ = \frac{2\sin 11A \sin A + 2\sin 7A \sin 3A}{2\cos 11A sin A + 2\cos 7A \sin 3A}\]
\[ = \frac{\cos \left( 11A - A \right) - \cos \left( 11A + A \right) + \cos \left( 7A - 3A \right) - \cos \left( 7A + 3A \right)}{\sin \left( 11A + A \right) - \sin \left( 11A - A \right) + \sin \left( 7A + 3A \right) - \sin \left( 7A - 3A \right)}\]
\[ = \frac{\cos 10A - \cos 12A + \cos 4A - \cos 10A}{\sin 12A - \sin 10A + \sin 10A - \sin 4A}\]
\[ = \frac{\cos 4A - \cos 12A}{\sin 12A - \sin 4A}\]
\[ = \frac{- 2\sin \left( \frac{4A + 12A}{2} \right) \sin \left( \frac{4A - 12A}{2} \right)}{2\sin \left( \frac{12A - 4A}{2} \right) \cos \left( \frac{12A + 4A}{2} \right)}\]
\[ = \frac{- \sin 8A \sin \left( - 4A \right)}{\sin 4A \cos 8A}\]
\[ = \frac{\sin 8A \sin 4A}{\sin 4A \cos 8A}\]
\[ = \tan8A\]
 = RHS
Hence, LHS = RHS.