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Question

$\frac{\text{c}-\text{b}\cos\text{A}}{\text{b}-\text{c}\cos\text{A}}=\frac{\cos\text{B}}{\cos\text{C}}$

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Answer

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

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