Let $A, B$ and $C$ be the angles of a triangle and $\tan \frac{A}{2}=\frac{2}{5}, \tan \frac{B}{2}=\frac{3}{7}$. Then, $\tan \frac{C}{2}$ is equal to

Let $$\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$$, where $$A, B$$ and $$C$$ are angles of a $$\triangle A B C$$. If the lengths of the sides opposite these angles are $$a, b$$ and $$c$$ respectively, then