In triangle ABC,with usual notations,2ac sin | vedclass

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

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

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(B) We know that in $	riangle ABC$,$A+B+C = \pi$,so $A+C = \pi-B$. Substituting this into the expression: $2ac \sin \left(\frac{(A+C)-B}{2}ight) = 2ac \sin \left(\frac{(\pi-B)-B}{2}ight)$ $= 2ac \sin \left(\frac{\pi-2B}{2}ight) = 2ac \sin \left(\frac{\pi}{2}-Bight)$ $= 2ac \cos B$ Using the cosine rule,$\cos B = \frac{a^2+c^2-b^2}{2ac}$. Substituting this value: $2ac \left(\frac{a^2+c^2-b^2}{2ac}ight) = a^2+c^2-b^2$.

In $\triangle ABC$,with usual notations,$2ac \sin \left(\frac{1}{2}(A-B+C)\right)$ is equal to

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