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

Answer

$\text{b}\cos\text{B + c}\cos\text{C}=\text{a}\cos(\text{B}-\text{C})$ Let $\text{a = k}\sin\text{A,b = k}\sin\text{B,c = k}\sin\text{C}$ $\text{LHS}=\text{b}\cos\text{B + c}\cos\text{C}$ $=\text{k}\sin\text{B}\cos\text{B}+\text{k}\sin\text{C}\cos\text{C}$ $=\frac{\text{k}}{2}(2\sin\text{B}\cos\text{B}+2\sin\text{C}\cos\text{C})$ $=\frac{\text{k}}{2}(\sin2\text{B}+\sin2\text{C})$ $=\frac{\text{k}}{2}2\sin(\text{B + C})\cos(\text{B}-\text{C})$ $=\text{k}\sin(\pi-\text{A})\cos(\text{B}-\text{C})$ $=\text{k}\sin\text{A}\cos(\text{B}-\text{C})$ $=\text{a}\cos(\text{B}-\text{C})=\text{RHS}$

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