Question

Prove that:

\[\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A\]

 

Answer 1

 Consider LHS: 
\[ \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}\]
\[ = \frac{\cos 4A + \cos 2A + \cos 3A}{\sin 4A + \sin 2A + \sin 3A}\]
\[ = \frac{2\cos \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \cos 3A}{2\sin \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \sin 3A}\]
\[ \]
\[ = \frac{2\cos 3A \cos A + \cos 3A}{2\sin 3A \cos A + \sin 3A}\]
\[ = \frac{\cos 3A\left[ 2\cos A + 1 \right]}{\sin 3A\left[ 2\cos A + 1 \right]}\]
\[ = \cot 3A\]
= RHS
Hence, RHS = LHS.