In a triangle ABC,with usual notations,2 ac s | vedclass

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

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

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

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

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