In ∆ABC, prove the following:
$c (a \cos B - b \cos A) = a^{2} - b^{2}$

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Solution

Consider

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

Hence proved.

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