In any triangle $A B C$ , prove the following
$c (a cos B - b c o s A) = a^{2} - b^{2}$

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Solution

$L H S = c (a cos B - b \times cos A)$
$== c ⎛ ⎜ ⎜ ⎝ a \frac{(c^{2} + a^{2} - b^{2})}{2 a c} - \frac{b (b^{2} + c^{2} - a^{2})}{2 b c} ⎞ ⎟ ⎟ ⎠$ (cosine rule)
$= \frac{c^{2} + a^{2} - b^{2} - b^{2} - c^{2} + a^{2}}{2}$
$= a^{2} - b^{2} = R H S$
$∴$ Hence, Proved

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