Question

If $\sin 2A=\lambda \sin 2B$, then write the value of $\frac{\lambda +1}{\lambda -1}$ ?

Answer 1

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

Applying componendo and dividendo

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

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

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