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In $\triangle \mathrm{ABC}$, with usual notations $\frac{\mathrm{b} \sin \mathrm{B}-\mathrm{c} \sin \mathrm{C}}{\sin (\mathrm{B}-\mathrm{C})}=$
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The correct answer is:
a
$\begin{aligned} & \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=\mathrm{k} \\ & \therefore \mathrm{a}=\mathrm{k} \sin \mathrm{A}, \mathrm{b}=\mathrm{k} \sin \mathrm{B}, \mathrm{c}=\mathrm{k} \sin \mathrm{C} \\ & \therefore \frac{\mathrm{b} \sin \mathrm{B}-\mathrm{c} \sin \mathrm{C}}{\sin (\mathrm{B}-\mathrm{C})}=\frac{\mathrm{k} \sin ^2 \mathrm{~B}-\mathrm{k} \sin ^2 \mathrm{C}}{\sin (\mathrm{B}-\mathrm{C})}=\frac{\mathrm{k}-\left[\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}\right]}{\sin ^{\mathrm{i}}(\mathrm{B}-\mathrm{C})} \\ & =\frac{\mathrm{k} \sin (\mathrm{B}-\mathrm{C}) \sin (\mathrm{B}+\mathrm{C})}{\sin (\mathrm{B}-\mathrm{C})} \\ & =\mathrm{k} \sin (\mathrm{B}+\mathrm{C}) \quad \mathrm{k}[\sin (\pi-\mathrm{A})]=\mathrm{k} \sin \mathrm{A}=\mathrm{a}\end{aligned}$
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