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

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(D) Given expression: $\sin 2A + \sin 2B - \sin 2C$Using the formula $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$:$= 2 \sin(A+B) \cos(A

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given expression: $\sin 2A + \sin 2B - \sin 2C$Using the formula $\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$:$= 2 \sin(A+B) \cos(A-B) - \sin 2C$Since $A+B+C = \pi$,we have $A+B = \pi - C$,so $\sin(A+B) = \sin C$ and $\cos(A+B) = -\cos C$.$= 2 \sin C \cos(A-B) - 2 \sin C \cos C$$= 2 \sin C [\cos(A-B) - \cos C]$$= 2 \sin C [\cos(A-B) + \cos(A+B)]$Using the formula $\cos(X-Y) + \cos(X+Y) = 2 \cos X \cos Y$:$= 2 \sin C [2 \cos A \cos B]$$= 4 \cos A \cos B \sin C$.Thus,the correct option is $(d)$.

If $A, B, C$ are angles of a triangle,then $\sin 2A + \sin 2B - \sin 2C$ is equal to

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