If \[ \sin2A=\lambda\sin2B \] prove that \[ \frac{\tan(A+B)}{\tan(A-B)} = \frac{\lambda+1}{\lambda-1} \]
Solution
Given:
\[ \sin2A=\lambda\sin2B \]Write \(\sin2A\) and \(\sin2B\) using identities:
\[ 2\sin A\cos A = \lambda(2\sin B\cos B) \]
\[
\sin A\cos A
=
\lambda\sin B\cos B
\]
Use identities:
\[ \sin2A+\sin2B = 2\sin(A+B)\cos(A-B) \] \[ \sin2A-\sin2B = 2\cos(A+B)\sin(A-B) \]Add and subtract the equations:
\[ \sin2A+\sin2B = (\lambda+1)\sin2B \] \[ \sin2A-\sin2B = (\lambda-1)\sin2B \]
\[
2\sin(A+B)\cos(A-B)
=
(\lambda+1)\sin2B
\]
\[
2\cos(A+B)\sin(A-B)
=
(\lambda-1)\sin2B
\]
Divide first equation by second equation:
\[ \frac{ 2\sin(A+B)\cos(A-B) }{ 2\cos(A+B)\sin(A-B) } = \frac{\lambda+1}{\lambda-1} \]
\[
\frac{\tan(A+B)}{\tan(A-B)}
=
\frac{\lambda+1}{\lambda-1}
\]
Hence Proved.