Prove that:
Solution
Consider LHS:
\[ \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}\]
Multiplying numerator and denominator by 2, we get
\[ = \frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}\]
\[ = \frac{\cos \left( A - 2A \right) - \cos \left( A + 2A \right) + \cos \left( 3A - 6A \right) - \cos \left( 3A + 6A \right)}{\sin \left( A + 2A \right) + \sin \left( A - 2A \right) + \sin \left( 3A + 6A \right) + \sin \left( 3A - 6A \right)}\]
\[ = \frac{\cos\left( - A \right) - \cos 3A + \cos \left( - 3A \right) - \cos 9A}{\sin 3A \sin\left( - A \right) + \sin 9A + \sin \left( - 3A \right)}\]
\[ = \frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A}\]
\[ = \frac{\cos A - \cos 9A}{\sin 9A - \sin A}\]
\[ = \frac{- 2\sin \left( \frac{A + 9A}{2} \right) \sin \left( \frac{A - 9A}{2} \right)}{2\cos \left( \frac{A + 9A}{2} \right) \sin \left( \frac{9A - A}{2} \right)}\]
\[ = \frac{\sin5A\cos4A}{\sin 5A \cos \left( - 4A \right)}\]
\[ = \tan 5A\]
= RHS
Hence, LHS = RHS.
