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

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(B) Given that $A, B, C$ are the angles of a triangle,so $A + B + C = \pi$. Using the identity $\sin 2A + \sin 2C = 2 \sin(A+C) \cos(A-C)$. Since $A+C

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$4 \cos A \sin B \cos C$

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(B) Given that $A, B, C$ are the angles of a triangle,so $A + B + C = \pi$. Using the identity $\sin 2A + \sin 2C = 2 \sin(A+C) \cos(A-C)$. Since $A+C = \pi - B$,we have $\sin(A+C) = \sin(\pi - B) = \sin B$. Thus,$\sin 2A + \sin 2C = 2 \sin B \cos(A-C)$. Now,the expression is: $\sin 2A - \sin 2B + \sin 2C = (\sin 2A + \sin 2C) - \sin 2B$ $= 2 \sin B \cos(A-C) - 2 \sin B \cos B$ $= 2 \sin B [\cos(A-C) - \cos B]$ Since $B = \pi - (A+C)$,$\cos B = \cos(\pi - (A+C)) = -\cos(A+C)$. $= 2 \sin B [\cos(A-C) + \cos(A+C)]$ Using $\cos(A-C) + \cos(A+C) = 2 \cos A \cos C$: $= 2 \sin B [2 \cos A \cos C] = 4 \cos A \sin B \cos C$.

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

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