If sin 2A=λ sin 2B,

Given:

sin 2A = λ sin 2B

$\Rightarrow \frac{\sin 2 A}{\sin 2 B}=\lambda$

$\Rightarrow \frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\lambda+1}{\lambda-1}$

$\Rightarrow \frac{2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)}{2 \sin \left(\frac{2 A-2 B}{2}\right) \cos \left(\frac{2 A+2 B}{2}\right)}=\frac{\lambda+1}{\lambda-1}$

$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right.$ and $\left.\sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$

$\Rightarrow \frac{\sin (A+B) \cos (A-B)}{\sin (A-B) \cos (A+B)}=\frac{\lambda+1}{\lambda-1}$

$\Rightarrow \tan (A+B) \cot (A-B)=\frac{\lambda+1}{\lambda-1}$

$\Rightarrow \frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$

Hence proved.