If sin (y+z-x), sin (z+x-y) and sin (x+y-z) a | vedclass

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(C) Given that $\sin (y+z-x), \sin (z+x-y), \sin (x+y-z)$ are in $AP$. Therefore,$2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$. Using the sum-to-prod

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$2 	an y=	an x+	an z$

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(C) Given that $\sin (y+z-x), \sin (z+x-y), \sin (x+y-z)$ are in $AP$. Therefore,$2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$. Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$,we get: $2 \sin (z+x-y) = 2 \sin \frac{(y+z-x) + (x+y-z)}{2} \cos \frac{(y+z-x) - (x+y-z)}{2}$ $2 \sin (z+x-y) = 2 \sin y \cos (z-x)$. Since $\sin (z+x-y) = \sin (z-(y-x)) = \sin z \cos (y-x) - \cos z \sin (y-x)$,this approach is complex. Alternatively,using $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$: $\sin (z+x-y) - \sin (y+z-x) = \sin (x+y-z) - \sin (z+x-y)$ $2 \cos z \sin (x-y) = 2 \cos x \sin (y-z)$ $\cos z (\sin x \cos y - \cos x \sin y) = \cos x (\sin y \cos z - \cos y \sin z)$ $\cos z \sin x \cos y - \cos z \cos x \sin y = \cos x \sin y \cos z - \cos x \cos y \sin z$ $2 \cos z \cos x \sin y = \cos x \cos y \sin z + \sin x \cos y \cos z$ Dividing by $\cos x \cos y \cos z$: $2 	an y = 	an z + 	an x$.

If $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $AP$,then

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