In $△ A B C$ , prove that, $a (b cos C - c cos B) = b^{2} - c^{2}$

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Solution

LHS $= a (b cos C - c cos B) = a b cos C - a c cos B$

$= a b \cdot \frac{a^{2} + b^{2} - c^{2}}{2 a b} - a c \cdot \frac{a^{2} + c^{2} - b^{2}}{2 a c}$ , ..... use cosine rule

$= \frac{1}{2} (a^{2} + b^{2} - c^{2} - a^{2} - c^{2} + b^{2}) = b^{2} - c^{2}$ =R.H.S.
Hence proved.

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