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

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(C) Given: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} = a^2 + b^2$ Using $\cos^2 \frac{C}{2} = \frac{1+\cos C}{2}$ and $\sin^2 \frac{C}{

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$\sin B$

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(C) Given: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} = a^2 + b^2$ Using $\cos^2 \frac{C}{2} = \frac{1+\cos C}{2}$ and $\sin^2 \frac{C}{2} = \frac{1-\cos C}{2}$: $\frac{(a-b)^2(1+\cos C)}{2} + \frac{(a+b)^2(1-\cos C)}{2} = a^2 + b^2$ $(a^2 + b^2 - 2ab)(1+\cos C) + (a^2 + b^2 + 2ab)(1-\cos C) = 2(a^2 + b^2)$ $(a^2 + b^2)(1+\cos C) - 2ab(1+\cos C) + (a^2 + b^2)(1-\cos C) + 2ab(1-\cos C) = 2(a^2 + b^2)$ $(a^2 + b^2)(1 + \cos C + 1 - \cos C) - 2ab(1 + \cos C - 1 + \cos C) = 2(a^2 + b^2)$ $2(a^2 + b^2) - 2ab(2 \cos C) = 2(a^2 + b^2)$ $-4ab \cos C = 0$ Since $a, b 
eq 0$,we have $\cos C = 0$,which implies $C = 90^{\circ}$. In $	riangle ABC$,$A + B + C = 180^{\circ}$,so $A + B = 90^{\circ}$,which means $A = 90^{\circ} - B$. Therefore,$\cos A = \cos(90^{\circ} - B) = \sin B$.

In a triangle $ABC$,if $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} = a^2 + b^2$,then $\cos A =$

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