Prove that: \[ \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A} {\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A \]
Solution
L.H.S.
\[ = \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A} {\cos 3A + \cos 5A + \cos 7A + \cos 9A} \]Group first & fourth terms and second & third terms.
Use identities:
\[ \sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
\frac{
2\sin6A\cos3A + 2\sin6A\cos A
}{
2\cos6A\cos3A + 2\cos6A\cos A
}
\]
Take common factor:
\[ = \frac{ 2\sin6A(\cos3A+\cos A) }{ 2\cos6A(\cos3A+\cos A) } \]Cancel common factors:
\[ = \frac{\sin6A}{\cos6A} \]Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ = \tan6A \]Hence Proved.