In any triangle `A B C ,` prove that: `(sin(B-C))/(sin(B+C))=(b^2-c^2)/(a^2)`

Explanation

`frac{sin(B-C)}{sin(B+C)}=frac{sinB.cosC-cosB.sinC}{sinB.cosC+cosB.sinC}`

By sine rule, `frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=k----(1)`

By cosine rule, `cos A=frac{b^2+c^2-a^2}{2bc}----(2)`

`cos B=frac{c^2+a^2-b^2}{2ca}----(3)`

`cos C=frac{a^2+b^2-c^2}{2ab}----(4)`

using (1), (2), (3) and (4),

`=frac{(frac{b}{k})(frac{a^2+b^2-c^2}{2ab})-(frac{a^2+c^2-b^2}{2ac})(frac{c}{k})}{(frac{b}{k})(frac{a^2+b^2-c^2}{2ab})+(frac{a^2+c^2-b^2}{2ac})(frac{c}{k})}`

`=frac{(frac{a^2+b^2-c^2}{2ak})-(frac{a^2+c^2-b^2}{2ak})}{(frac{a^2+b^2-c^2}{2ak})+(frac{a^2+c^2-b^2}{2ak})}`
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