In a triangle ABC,if (b+c)^2 sin^2left(frac{A | vedclass

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(B) $L.H.S. = (b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right)$$= (b^2 + c^2 + 2bc) \sin^2\left(\frac{A}{2}\right) + (b

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$2R^2$

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(B) $L.H.S. = (b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right)$$= (b^2 + c^2 + 2bc) \sin^2\left(\frac{A}{2}\right) + (b^2 + c^2 - 2bc) \cos^2\left(\frac{A}{2}\right)$$= (b^2 + c^2) \left[\sin^2\left(\frac{A}{2}\right) + \cos^2\left(\frac{A}{2}\right)\right] - 2bc \left[\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)\right]$$= b^2 + c^2 - 2bc \cos A$$= a^2$ (Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$)Since $a = 2R \sin A$,we have $a^2 = 4R^2 \sin^2 A$$= 4R^2 \left(\frac{1 - \cos 2A}{2}\right)$$= 2R^2 (1 - \cos 2A)$Comparing with $K(1 - \cos 2A)$,we get $K = 2R^2$.

In a triangle $ABC$,if $(b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right) = K(1 - \cos 2A)$,then $K =$

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