In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $. In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.
Please help. Thanks in advance.

Note that
\begin{align*}
\frac{\sin(B-C)}{\sin (B+C)}&=\frac{\sin B \cos C-\cos B \sin C}{\sin B \cos C+\cos B \sin C}\\
&=\frac{\frac{\sin B}{\sin C} \cos C-\cos B}{\frac{\sin B}{\sin C} \cos C+\cos B}\\
&=\frac{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}-\frac{b^2-a^2-c^2}{2ac}}{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}+\frac{b^2-a^2-c^2}{2ac}}\\
&=\frac{b^2-c^2}{a^2}.
\end{align*}
