Prove that: \[ \frac{ \sin11A\sin A+\sin7A\sin3A }{ \cos11A\sin A+\cos7A\sin3A } = \tan8A \]
Solution
L.H.S.
\[ = \frac{ \sin11A\sin A+\sin7A\sin3A }{ \cos11A\sin A+\cos7A\sin3A } \]Use identities:
\[ \sin C\sin D = \frac12[\cos(C-D)-\cos(C+D)] \] \[ \cos C\sin D = \frac12[\sin(C+D)-\sin(C-D)] \]Applying identities in numerator:
\[ \sin11A\sin A = \frac12(\cos10A-\cos12A) \] \[ \sin7A\sin3A = \frac12(\cos4A-\cos10A) \]
\[
=
\frac{
\frac12(\cos10A-\cos12A)
+
\frac12(\cos4A-\cos10A)
}{
\cos11A\sin A+\cos7A\sin3A
}
\]
Simplify numerator:
\[ = \frac{ \frac12(\cos4A-\cos12A) }{ \cos11A\sin A+\cos7A\sin3A } \]Use identity:
\[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ \cos4A-\cos12A = 2\sin8A\sin4A \] Hence numerator becomes: \[ = \sin8A\sin4A \]Applying identity in denominator:
\[ \cos11A\sin A = \frac12(\sin12A-\sin10A) \] \[ \cos7A\sin3A = \frac12(\sin10A-\sin4A) \]
\[
=
\frac{
\sin8A\sin4A
}{
\frac12(\sin12A-\sin4A)
}
\]
Use identity:
\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ \sin12A-\sin4A = 2\cos8A\sin4A \] Hence denominator becomes: \[ = \cos8A\sin4A \]
\[
=
\frac{\sin8A\sin4A}{\cos8A\sin4A}
=
\frac{\sin8A}{\cos8A}
\]
Use identity:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ = \tan8A \]Hence Proved.