In a Delta ABC, 2ac sin left( frac{A - B + C} | vedclass

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(B) We know that in a $\Delta ABC,$ $A + B + C = \pi,$ so $A + C = \pi - B.$Substituting this into the expression: $\frac{A - B + C}{2} = \frac{(A + C

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$c^2 + a^2 - b^2$

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(B) We know that in a $\Delta ABC,$ $A + B + C = \pi,$ so $A + C = \pi - B.$Substituting this into the expression: $\frac{A - B + C}{2} = \frac{(A + C) - B}{2} = \frac{(\pi - B) - B}{2} = \frac{\pi - 2B}{2} = \frac{\pi}{2} - B.$Thus,$2ac \sin \left( \frac{\pi}{2} - B \right) = 2ac \cos B.$Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$Therefore,$2ac \cos B = 2ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = a^2 + c^2 - b^2.$

In a $\Delta ABC,$ $2ac \sin \left( \frac{A - B + C}{2} \right)$ is equal to

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