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

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

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

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(B) We know that in a $	riangle ABC$,$A+B+C = \pi$,so $A+C = \pi - B$.Substituting this into the expression,we get $\frac{A+C-B}{2} = \frac{\pi-B-B}{2} = \frac{\pi}{2} - B$.Thus,$2ac \sin \left(\frac{A-B+C}{2}ight) = 2ac \sin \left(\frac{\pi}{2}-Bight)$.Using the identity $\sin \left(\frac{\pi}{2}-	hetaight) = \cos 	heta$,we get $2ac \cos B$.From the Law of Cosines,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.Substituting this,we get $2ac \left(\frac{a^2+c^2-b^2}{2ac}ight) = a^2+c^2-b^2$.

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

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