In any triangle ABC,a(b cos C - c cos B) equa | vedclass

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(B) Using the projection formula or the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting t

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$b^2 - c^2$

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(B) Using the projection formula or the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting these into the expression: $a(b \cos C - c \cos B) = ab \cos C - ac \cos B$ $= ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right)$ $= \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2}$ $= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$ $= \frac{2b^2 - 2c^2}{2}$ $= b^2 - c^2$.

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

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