Question
$\text{b}(\text{c}\cos\text{A}-\text{a}\cos\text{C})=\text{c}^2-\text{a}^2$
$\text{RHS}=\text{c}^2-\text{a}^2$
$=\text{k}^2\sin^2\text{C}-\text{k}^2\sin^2\text{A}$
$=\text{k}^2(\sin^2\text{C}-\sin^2\text{A})$
$=\text{k}^2\sin(\text{C + A}).\sin(\text{C}-\text{A})$
$=\text{k}^2\sin(\pi-\text{B}).\sin(\text{C}-\text{A})$
$=\text{k}^2\sin\text{B}.\sin(\text{C}-\text{A})$
$=\text{k}\sin\text{B}.\text{k}\sin(\text{C}-\text{A})$
$=\text{bk}\sin(\text{C}-\text{A})$
$=\text{bk}(\sin\text{C}.\cos\text{A}-\sin\text{A}.\cos\text{C})$
$=\text{b}(\text{k}\sin\text{C}.\cos\text{A}-\text{k}\sin\text{A}.\cos\text{C})$
$=\text{b}(\text{c}\cos\text{A}-\text{a}\cos\text{C})=\text{LHS}$
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