Prove that: \[ \frac{ \sin3A\cos4A-\sin A\cos2A }{ \sin4A\sin A+\cos6A\cos A } = \tan2A \]
Solution
L.H.S.
\[ = \frac{ \sin3A\cos4A-\sin A\cos2A }{ \sin4A\sin A+\cos6A\cos A } \]Use identities:
\[ \sin C\cos D = \frac12[\sin(C+D)+\sin(C-D)] \] \[ \sin C\sin D = \frac12[\cos(C-D)-\cos(C+D)] \] \[ \cos C\cos D = \frac12[\cos(C-D)+\cos(C+D)] \]Applying identity in numerator:
\[ \sin3A\cos4A = \frac12(\sin7A-\sin A) \] \[ \sin A\cos2A = \frac12(\sin3A-\sin A) \]
\[
=
\frac{
\frac12(\sin7A-\sin A)-\frac12(\sin3A-\sin A)
}{
\sin4A\sin A+\cos6A\cos A
}
\]
Simplify numerator:
\[ = \frac{ \frac12(\sin7A-\sin3A) }{ \sin4A\sin A+\cos6A\cos A } \]Use identity:
\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ \sin7A-\sin3A = 2\cos5A\sin2A \] Hence numerator becomes: \[ = \cos5A\sin2A \]Applying identities in denominator:
\[ \sin4A\sin A = \frac12(\cos3A-\cos5A) \] \[ \cos6A\cos A = \frac12(\cos5A+\cos7A) \]
\[
=
\frac{
\cos5A\sin2A
}{
\frac12(\cos3A+\cos7A)
}
\]
Use identity:
\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \cos3A+\cos7A = 2\cos5A\cos2A \] Hence denominator becomes: \[ = \cos5A\cos2A \]
\[
=
\frac{\cos5A\sin2A}{\cos5A\cos2A}
=
\frac{\sin2A}{\cos2A}
\]
Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ = \tan2A \]Hence Proved.