In any triangle ABC,with usual notations,c(a | vedclass

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(A) Using the projection formula or the law of cosines:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Substituting t

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$a^2 - b^2$

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(A) Using the projection formula or the law of cosines:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Substituting these into the expression:$c(a \cos B - b \cos A) = c \left( a \left( \frac{c^2 + a^2 - b^2}{2ac} \right) - b \left( \frac{b^2 + c^2 - a^2}{2bc} \right) \right)$$= c \left( \frac{c^2 + a^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c} \right)$$= \frac{c^2 + a^2 - b^2 - b^2 - c^2 + a^2}{2}$$= \frac{2a^2 - 2b^2}{2} = a^2 - b^2$

In any $\triangle ABC$,with usual notations,$c(a \cos B - b \cos A) =$

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