Question

If $\sin 2A=\lambda \sin 2B$, prove that $\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda +1}{\lambda -1}$

Answer 1

$\sin 2A=\lambda \sin 2B$

RHS

$=\frac{\lambda +1}{\lambda -1}$

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

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

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

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

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

=LHS

Hence proved.