In Delta ABC, a(bcos C - ccos B) = | vedclass

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(B) Using the projection formula,we know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting these val

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$b^2 - c^2$

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(B) Using the projection formula,we know that $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting these values into the expression:$a(b\cos C - c\cos B) = a \left( b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right)$$= a \left( \frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right)$$= \frac{a^2 + b^2 - c^2 - (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 $\Delta ABC, a(b\cos C - c\cos B) = $

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