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

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(C) Given that $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A$.$P$.$	herefore 2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$Using the formu

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$2 	an y = 	an x + 	an z$

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(C) Given that $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A$.$P$.$	herefore 2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$Using the formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:$2 \sin (z+x-y) = 2 \sin \left( \frac{y+z-x+x+y-z}{2} ight) \cos \left( \frac{y+z-x-x-y+z}{2} ight)$$2 \sin (z+x-y) = 2 \sin y \cos (z-x)$Dividing both sides by $\cos (y+z-x) \cos (z+x-y) \cos (x+y-z)$ is not direct,so we use the property $\frac{\sin (A-B)}{\cos A \cos B} = 	an A - 	an B$.By simplifying the $A$.$P$. condition,we get $	an x, 	an y, 	an z$ are in $A$.$P$.Therefore,$2 	an y = 	an x + 	an z$.

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

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