If \( \sin2A=\lambda\sin2B \), then write the value of \( \dfrac{\lambda+1}{\lambda-1} \)
Solution:
Given,
\[
\sin2A=\lambda\sin2B
\]
\[
\lambda=\frac{\sin2A}{\sin2B}
\]
Therefore,
\[
\frac{\lambda+1}{\lambda-1}
=
\frac{
\frac{\sin2A}{\sin2B}+1
}{
\frac{\sin2A}{\sin2B}-1
}
\]
\[
=
\frac{\sin2A+\sin2B}{\sin2A-\sin2B}
\]
Using identities,
\[
\sin C+\sin D
=
2\sin\frac{C+D}{2}\cos\frac{C-D}{2}
\]
and
\[
\sin C-\sin D
=
2\cos\frac{C+D}{2}\sin\frac{C-D}{2}
\]
\[
=
\frac{
2\sin(A+B)\cos(A-B)
}{
2\cos(A+B)\sin(A-B)
}
\]
\[
=
\tan(A+B)\cot(A-B)
\]
\[
\boxed{\tan(A+B)\cot(A-B)}
\]