Prove that: \[ \frac{ \sin 5A \cos 2A – \sin 6A \cos A }{ \sin A \sin 2A – \cos 2A \cos 3A } = \tan A \]
Solution
L.H.S.
\[ = \frac{ \sin 5A \cos 2A – \sin 6A \cos A }{ \sin A \sin 2A – \cos 2A \cos 3A } \]Use identity:
\[ \sin C \cos D = \frac{1}{2} [\sin(C+D)+\sin(C-D)] \]Applying identity in numerator:
\[ \sin5A\cos2A = \frac{1}{2}(\sin7A+\sin3A) \] \[ \sin6A\cos A = \frac{1}{2}(\sin7A+\sin5A) \]
\[
=
\frac{
\frac12(\sin7A+\sin3A)-\frac12(\sin7A+\sin5A)
}{
\sin A \sin2A-\cos2A\cos3A
}
\]
Simplify numerator:
\[ = \frac{ \frac12(\sin3A-\sin5A) }{ \sin A \sin2A-\cos2A\cos3A } \]Use identity:
\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
\sin3A-\sin5A
=
2\cos4A\sin(-A)
=
-2\cos4A\sin A
\]
Hence numerator becomes:
\[
=-\cos4A\sin A
\]
Use identity in denominator:
\[ \cos C\cos D-\sin C\sin D = \cos(C+D) \] \[ \sin A\sin2A-\cos2A\cos3A = -\cos(A+2A) = -\cos A \]
\[
=
\frac{-\cos A\sin A}{-\cos^2A}
=
\frac{\sin A}{\cos A}
\]
Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ = \tan A \]Hence Proved.