In a triangle ABC,a[b cos C - c cos B] = | vedclass

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(D) We are given the expression $a[b \cos C - c \cos B]$.Using the projection formula for a triangle,we know that $a = b \cos C + c \cos B$.Substituti

Vedclass · Accepted Answer

$b^2 - c^2$

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(D) We are given the expression $a[b \cos C - c \cos B]$.Using the projection formula for a triangle,we know that $a = b \cos C + c \cos B$.Substituting this into the expression,we get $(b \cos C + c \cos B)(b \cos C - c \cos B)$.This is in the form $(x+y)(x-y) = x^2 - y^2$,so it simplifies to $b^2 \cos^2 C - c^2 \cos^2 B$.Alternatively,using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting these values: $a \left[ b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right]$.$= a \left[ \frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right]$.$= a \left[ \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2a} \right]$.$= a \left[ \frac{2b^2 - 2c^2}{2a} \right] = b^2 - c^2$.

In a triangle $ABC$,$a[b \cos C - c \cos B] = $

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