Question
$\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})=0$

Answer

$\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})=0$ $\text{LHS}=\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})$ $=\text{a}\sin\text{B}-\text{a}\sin\text{C}+\text{b}\sin\text{C}-\text{b}\sin\text{A}+\text{c}\sin\text{A}-\text{c}\sin\text{B}$ $=\text{b}\sin\text{A}-\text{c}\sin\text{A + c}\sin\text{B}-\text{b}\sin\text{A + c}\sin\text{A}-\text{c}\sin\text{B}$ $[\because \text{b}\sin\text{A = a}\sin\text{B,b}\sin\text{C = c}\sin]$ $=0=\text{RHS}$ Hence Proved

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