Prove that: \[ \frac{\cos 4A + \cos 3A + \cos 2A} {\sin 4A + \sin 3A + \sin 2A} = \cot 3A \]
Solution
L.H.S.
\[ = \frac{\cos 4A + \cos 3A + \cos 2A} {\sin 4A + \sin 3A + \sin 2A} \]Group first and third terms and use identities:
\[ \cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
\frac{
2\cos3A\cos A + \cos3A
}{
2\sin3A\cos A + \sin3A
}
\]
Take common factor:
\[ = \frac{ \cos3A(2\cos A + 1) }{ \sin3A(2\cos A + 1) } \]Cancel common factor:
\[ = \frac{\cos3A}{\sin3A} \]Use identity:
\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \] \[ = \cot3A \]Hence Proved.