In a triangle ABC,(a-b)^2 cos^2 frac{C}{2} + | vedclass

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(B) We have,\ $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ \ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C

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$c^2$

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(B) We have,\ $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ \ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$ \ $= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ \ $= (a^2 + b^2)(1) - 2ab \cos C$ \ $= a^2 + b^2 - 2ab \cos C$ \ Since by the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,\ Therefore,the expression is equal to $c^2$.

In a $\triangle ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to

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