$\begin{align}  & \text{In a triangle }ABC \\ & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=K \\ & K\sin C=c---(i) \\ & K\sin B=b---(ii) \\ & \text{Add (i) and (ii) together} \\ & K\sin B+K\sin C=b+c---(iii) \\ & \text{Subtract (i) from (ii)} \\ & K\sin B-K\sin C=b-c---(iv) \\ & \text{Divide (iii) by (iv)} \\ & \frac{K(\sin B+\sin C)}{K(\sin B-\sin C)}=\frac{b+c}{b-c} \\ & \frac{\sin B+\sin C}{\sin B-\sin C}=\frac{b+c}{b-c} \\ & \frac{2\sin \tfrac{B+C}{2}\cos \tfrac{B-C}{2}}{2\cos \tfrac{B+C}{2}\sin \tfrac{B-C}{2}}=\frac{b+c}{b-c} \\ & \tan \tfrac{B+C}{2}\cot \tfrac{B-C}{2}=\frac{b+c}{b-c} \\ & \text{Note: }{{180}^{\circ }}-A=B+C \\ & \tan \left( \tfrac{{{180}^{\circ }}-A}{2} \right)\cot \tfrac{B-C}{2}=\frac{b+c}{b-c} \\ & \tan \left( {{90}^{\circ }}-\frac{A}{2} \right)\cot \tfrac{B-C}{2}=\frac{b+c}{b-c} \\ & \cot \tfrac{A}{2}\cot \tfrac{B-C}{2}=\frac{b+c}{b-c} \\ & \cot \tfrac{A}{2}=\left( \frac{1}{\cot \tfrac{B-C}{2}} \right)\frac{b+c}{b-c} \\ & \cot \tfrac{A}{2}=\tan \left( \frac{B-C}{2} \right)\frac{b+c}{b-c} \\ & \tan \left( \frac{B-C}{2} \right)=\frac{b-c}{b+c}\cot \frac{A}{2} \\\end{align}$