Question
$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$

Answer

$\text{b}\sin\text{B}-\text{c}\sin\text{C = a}\sin(\text{B}-\text{C})$ $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$ $\text{RHS}=\text{a}\sin(\text{B}-\text{C})$ $=\text{a}\sin\text{B}.\cos\text{C}-\text{a}\sin \text{C}.\cos\text{B}$ $=\text{a(bk).}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{a(ck)}.\Big(\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}\Big)$ $=\text{k}.\frac{(\text{a}^2+\text{b}^2-\text{c}^2)}{2}-\text{k}\frac{(\text{a}^2+\text{c}^2-\text{b}^2)}{2}$ $=2\text{k}.\frac{(\text{b}^2-\text{c}^2)}{2}$ $=\text{b.(kb)}-\text{c(kc)}$ $=\text{b}(\sin\text{B})-\text{c}(\sin\text{C})=\text{LHS}$

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