In triangle ABC,(a-b)^2 sin^2left(frac{A+B}{2 | vedclass

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(C) In $	riangle ABC$,$A+B+C = \pi$,so $\frac{A+B}{2} = \frac{\pi-C}{2}$.Substituting this into the expression:$(a-b)^2 \sin^2\left(\frac{\pi-C}{2}

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$c^2$

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(C) In $	riangle ABC$,$A+B+C = \pi$,so $\frac{A+B}{2} = \frac{\pi-C}{2}$.Substituting this into the expression:$(a-b)^2 \sin^2\left(\frac{\pi-C}{2}ight) + (a+b)^2 \sin^2\left(\frac{C}{2}ight)$$= (a-b)^2 \cos^2\left(\frac{C}{2}ight) + (a+b)^2 \sin^2\left(\frac{C}{2}ight)$$= (a^2+b^2-2ab) \cos^2\left(\frac{C}{2}ight) + (a^2+b^2+2ab) \sin^2\left(\frac{C}{2}ight)$$= (a^2+b^2) \left(\cos^2\frac{C}{2} + \sin^2\frac{C}{2}ight) - 2ab \left(\cos^2\frac{C}{2} - \sin^2\frac{C}{2}ight)$$= (a^2+b^2)(1) - 2ab \cos C$Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$:$= a^2+b^2 - 2ab \left(\frac{a^2+b^2-c^2}{2ab}ight)$$= a^2+b^2 - (a^2+b^2-c^2) = c^2$.

In $\triangle ABC$,$(a-b)^2 \sin^2\left(\frac{A+B}{2}\right) + (a+b)^2 \sin^2\left(\frac{C}{2}\right) = $

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