$\begin{align}  & \text{The sine rule state that } \\ & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \\ & b=2R\sin B,\text{ }a=2R\sin A,\text{ }c=2R\sin C \\ & \frac{b-c}{a}=\frac{2R\sin B-2R\sin C}{2R\sin A}=\frac{2R(\sin B-\sin C)}{2R\sin A} \\ & \frac{b-c}{a}=\frac{\sin B-\sin C}{\sin A} \\ & \frac{b-c}{a}=\frac{2\cos \tfrac{B+C}{2}\sin \tfrac{B-C}{2}}{\sin [180-(B+C)]}=\frac{2\cos \tfrac{B+C}{2}\sin \tfrac{B-C}{2}}{\sin (B+C)} \\ & \frac{b-c}{a}=\frac{2\cos \tfrac{B+C}{2}\sin \tfrac{B-C}{2}}{2\sin \tfrac{B+C}{2}\cos \tfrac{B+C}{2}}=\frac{\sin \tfrac{B-C}{2}}{\sin \tfrac{B+C}{2}} \\ & \frac{b-c}{2}=\frac{\sin \tfrac{B-C}{2}}{\sin (\tfrac{180-A}{2})}=\frac{\sin \tfrac{B-C}{2}}{\sin (90-\tfrac{A}{2})} \\ & \frac{b-c}{a}=\frac{\sin \tfrac{B-C}{2}}{\cos \tfrac{A}{2}} \\\end{align}$