Prove that: \[ \frac{ \sin 5A – \sin 7A + \sin 8A – \sin 4A }{ \cos 4A + \cos 7A – \cos 5A – \cos 8A } = \cot 6A \]
Solution
L.H.S.
\[ = \frac{ \sin 5A – \sin 7A + \sin 8A – \sin 4A }{ \cos 4A + \cos 7A – \cos 5A – \cos 8A } \]Group terms and use identities:
\[ \sin C – \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ \cos C – \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
=
\frac{
2\cos6A\sin(-A)
+
2\cos6A\sin2A
}{
-2\sin\frac{9A}{2}\sin\left(-\frac{A}{2}\right)
–
2\sin\frac{13A}{2}\sin\left(-\frac{3A}{2}\right)
}
\]
Simplifying signs:
\[ = \frac{ -2\cos6A\sin A + 2\cos6A\sin2A }{ 2\sin\frac{9A}{2}\sin\frac{A}{2} + 2\sin\frac{13A}{2}\sin\frac{3A}{2} } \]After simplification and taking common factors:
\[ = \frac{\cos6A}{\sin6A} \]Use identity:
\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \] \[ = \cot6A \]Hence Proved.