Question
$\text{c}(\text{a}\cos\text{B}-\text{b}\cos\text{A})=\text{a}^2-\text{b}^2$

Answer

$\text{c}(\text{a}\cos\text{B}-\text{b}\cos\text{A})$
$=\text{ac}.\cos\text{B}-\text{bc}\cos\text{A}$
$=\text{ac}.\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}-\text{bc}.\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2}-\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2}$
$=\frac{\text{a}^2+\text{c}^2-\text{b}^2-\text{b}^2-\text{c}^2+\text{a}^2}{2}$
$=\frac{2\text{a}^2-2\text{b}^2}{2}=(\text{a}^2-\text{b}^2)$

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