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.


In the above configuration, $R \sin(B-C) = OJ_A = M_A H_A$ and $R\sin(B+C)=R\sin(A)= BM_A$. Moreover, by the Pythagorean theorem we have $H_A B^2-H_A C^2=AB^2-AC^2=c^2-b^2$, hence
$ b^2-c^2 = (H_A B+H_A C)(H_A B-H_A C) $ and
$ H_A B - H_A C = \frac{c^2-b^2}{a}.$
It follows that
$$ BH_A = \frac{1}{2}\left(\frac{c^2-b^2}{a}+a\right) = \frac{a^2+c^2-b^2}{2a} $$
and
$$ M_A H_A = \frac{a}{2}-BH_A = \frac{b^2-c^2}{2a} $$
so
$$ \boxed{\frac{\sin(B-C)}{\sin(B+C)}=\frac{OJ_A}{BM_A} = \frac{2 M_A H_A}{a} = \color{red}{\frac{b^2-c^2}{a^2}}.}$$

An alternative approach through the sine addition formula and the cosine theorem:
$$\begin{eqnarray*}\frac{\sin(B-C)}{\sin(B+C)}=\frac{2R \sin(B-C)}{2R\sin(A)}&=&\frac{2R\sin(B)\cos(C)-2R\sin(C)\cos(B)}{a}\\&=&\frac{2ab\cos(C)-2ac\cos(B)}{2a^2}\\&=&\frac{(a^2+b^2-c^2)-(a^2+c^2-b^2)}{2a^2}\\&=&\color{red}{\frac{b^2-c^2}{a^2}}.\end{eqnarray*}$$
