If sin A + sin B = sqrt{3}(cos B - cos A),the | vedclass

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(A) Given,$\sin A + \sin B = \sqrt{3}(\cos B - \cos A)$.Rearranging the terms,we get $\sin A + \sqrt{3} \cos A = \sqrt{3} \cos B - \sin B$.Dividing bo

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(A) Given,$\sin A + \sin B = \sqrt{3}(\cos B - \cos A)$.Rearranging the terms,we get $\sin A + \sqrt{3} \cos A = \sqrt{3} \cos B - \sin B$.Dividing both sides by $2$,we get $\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A = \frac{\sqrt{3}}{2} \cos B - \frac{1}{2} \sin B$.Using the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$,we have:$\sin(A + \frac{\pi}{3}) = \sin(\frac{\pi}{3} - B)$.This implies $A + \frac{\pi}{3} = \frac{\pi}{3} - B$,which simplifies to $A = -B$ or $A + B = 0$.Now,we need to find the value of $\sin 3A + \sin 3B$.Since $A = -B$,then $3A = -3B$.Therefore,$\sin 3A + \sin 3B = \sin(-3B) + \sin 3B = -\sin 3B + \sin 3B = 0$.

If $\sin A + \sin B = \sqrt{3}(\cos B - \cos A)$,then $\sin 3A + \sin 3B$ is equal to:

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