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

We have
\(2 \sin (z+x-y)=\sin (y+z-x)+\sin (x+y-z)\)
\( \therefore \sin (\mathrm{y}+\mathrm{z}-\mathrm{x})-\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})=\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})\)\(-\sin (\mathrm{x}+\mathrm{y}-\mathrm{z}) \)
\( \therefore 2 \cos \mathrm{z} \sin (\mathrm{y}-\mathrm{x})=2 \cos \mathrm{x} \sin (\mathrm{z}-\mathrm{y}) \)
\( \therefore \cos \mathrm{z}(\sin \mathrm{y} \cos \mathrm{x}-\cos \mathrm{y} \sin \mathrm{x})=\cos \mathrm{x}(\sin \mathrm{z} \cos \mathrm{y}\)\(-\cos \mathrm{z} \sin \mathrm{y}) \)
\( \therefore \cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z}-\sin \mathrm{x} \cos \mathrm{y} \cos \mathrm{z}=\cos \mathrm{x} \cos \mathrm{y} \sin \mathrm{z} \)
\( -\cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z} \)
\( \text { Dividing both } \operatorname{sides} \text { by } \cos \mathrm{x} \cos \mathrm{y} \cos \mathrm{z} \text {, we get } \)
\( \tan \mathrm{y}-\tan \mathrm{x}=\tan \mathrm{z}-\tan \mathrm{y} \)
\( \therefore 2 \tan \mathrm{y}=\tan \mathrm{x}+\tan \mathrm{z}\)