Question

Prove that:

\[\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A\]

Answer 1

Consider LHS: 
\[ \frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}\]
Multiplying numerator and denominator by 2, we get
\[ = \frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}\]
\[ = \frac{\sin \left( 5A + 2A \right) + \sin \left( 5A - 2A \right) - \sin \left( 6A + A \right) - \sin \left( 6A - A \right)}{\cos \left( A - 2A \right) + \cos \left( A + 2A \right) - \cos \left( 2A + 3A \right) - \cos \left( 2A - 3A \right)}\]
\[ = \frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos \left( - A \right) + \cos 3A - \cos 5A - \cos \left( - A \right)}\]
\[ = \frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos A + \cos 3A - \cos 5A - cos A}\]
\[ = \frac{\sin 3A - \sin 5A}{\cos 3A - \cos 5A}\]
\[ = \frac{2\sin \left( \frac{3A - 5A}{2} \right) \cos \left( \frac{3A + 5A}{2} \right)}{- 2\cos \left( \frac{3A + 5A}{2} \right) \cos\left( \frac{3A - 5A}{2} \right)}\]
\[ = \frac{\sin \left( - A \right) \cos 4A}{- \cos 4A \cos \left( - A \right)}\]
\[ = \frac{- \sin A \cos 4A}{- \cos 4A\cos A}\]
\[ = \frac{\sin A}{\cos A}\]
\[ = \tan A\]
 = RHS
Hence, LHS = RHS .