If A, B, C are the angles of a triangle,then | vedclass

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(B) In a triangle $ABC$,$A + B + C = \pi$,so $C = \pi - (A + B)$.We know that $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$.Proof:$\sin

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$2$

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(B) In a triangle $ABC$,$A + B + C = \pi$,so $C = \pi - (A + B)$.We know that $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$.Proof:$\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 A + \sin^2 B + \sin^2 (A + B)$$= \sin^2 A + \sin^2 B + (\sin A \cos B + \cos A \sin B)^2$$= \sin^2 A + \sin^2 B + \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos B \cos A \sin B$$= \sin^2 A(1 + \cos^2 B) + \sin^2 B(1 + \cos^2 A) + 2 \sin A \cos A \sin B \cos B$Alternatively,using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A \cos B \cos C$ for any triangle $ABC$:$\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = (2 + 2\cos A \cos B \cos C) - 2\cos A \cos B \cos C = 2$.Thus,the correct option is $B$.

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

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