Question

Let $\sin A=m\sin B$ then prove that $\tan \frac{A-B}{2}=\frac{m-1}{m+1}\tan \frac{A+B}{2}$.

Answer 1

Now,

$\sin A=m\sin B$

or, $\frac{\sin A}{\sin B}=m$

or, $\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{m-1}{m+1}$

or, $\frac{2\sin \frac{A-B}{2}\cos \frac{A+B}{2}}{2\sin \frac{A+B}{2}.\cos \frac{A-B}{2}}=\frac{m-1}{m+1}$

or, $\tan \frac{A-B}{2}=\frac{m-1}{m+1}\tan \frac{A+B}{2}$.