Let $\Delta ABC(\angle B > \angle C)$ and $O,I,O_1$ are respectively the centre of circumcircle, Incircle and the centre of the excircle of $\Delta ABC$ touching the side $BC$. Prove that $$\frac{\tan \angle IOO_1}{2}=\frac{\sin B-\sin C}{2\cos A-1}$$
What all i tried $$\frac{\tan \angle IOO_1}{2}=\frac{\frac{\sin \angle IOO_1}{\cos \angle IOO_1}}{2}=\frac{\frac{\sin \angle IOO_1}{\frac{OI^2+OO_1^2-IO_1^2}{2OI\cdot OO_1}}}{2}=\frac{\sin \angle IOO_1}{\frac{OI^2+OO_1^2-IO_1^2}{OI\cdot OO_1}}$$
Then by Euler formula but i am stuck. Help