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If $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A P$, then
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Verified Answer
The correct answer is:
$2 \tan y=\tan x+\tan z$
We have $2 \sin (z+x-y)=\sin (y+z-x)+\sin (x+y-z)$
$$
\begin{aligned}
& \therefore \sin (\mathrm{y}+\mathrm{z}-\mathrm{x})-\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})=\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})-\sin (\mathrm{x}+\mathrm{y}-\mathrm{z}) \\
& \therefore 2 \cos \mathrm{z} \sin (\mathrm{y}-\mathrm{x})=2 \cos \mathrm{x} \sin (\mathrm{z}-\mathrm{y}) \\
& \therefore \cos \mathrm{z}(\sin \mathrm{y} \cos \mathrm{x}-\cos \mathrm{y} \sin \mathrm{x})=\cos \mathrm{x}(\sin \mathrm{z} \cos \mathrm{y}-\cos \mathrm{z} \sin \mathrm{y}) \\
& \therefore \cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z}-\sin \mathrm{x} \cos \mathrm{y} \cos \mathrm{z}=\cos \mathrm{x} \cos \mathrm{y} \sin \mathrm{z} \\
& -\cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z} \\
& \text { Dividing both } \operatorname{sides} \text { by } \cos \mathrm{x} \cos \mathrm{y} \cos \mathrm{z} \text {, we get } \\
& \tan \mathrm{y}-\tan \mathrm{x}=\tan \mathrm{z}-\tan \mathrm{y} \\
& \therefore 2 \tan \mathrm{y}=\tan \mathrm{x}+\tan \mathrm{z}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \sin (\mathrm{y}+\mathrm{z}-\mathrm{x})-\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})=\sin (\mathrm{z}+\mathrm{x}-\mathrm{y})-\sin (\mathrm{x}+\mathrm{y}-\mathrm{z}) \\
& \therefore 2 \cos \mathrm{z} \sin (\mathrm{y}-\mathrm{x})=2 \cos \mathrm{x} \sin (\mathrm{z}-\mathrm{y}) \\
& \therefore \cos \mathrm{z}(\sin \mathrm{y} \cos \mathrm{x}-\cos \mathrm{y} \sin \mathrm{x})=\cos \mathrm{x}(\sin \mathrm{z} \cos \mathrm{y}-\cos \mathrm{z} \sin \mathrm{y}) \\
& \therefore \cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z}-\sin \mathrm{x} \cos \mathrm{y} \cos \mathrm{z}=\cos \mathrm{x} \cos \mathrm{y} \sin \mathrm{z} \\
& -\cos \mathrm{x} \sin \mathrm{y} \cos \mathrm{z} \\
& \text { Dividing both } \operatorname{sides} \text { by } \cos \mathrm{x} \cos \mathrm{y} \cos \mathrm{z} \text {, we get } \\
& \tan \mathrm{y}-\tan \mathrm{x}=\tan \mathrm{z}-\tan \mathrm{y} \\
& \therefore 2 \tan \mathrm{y}=\tan \mathrm{x}+\tan \mathrm{z}
\end{aligned}
$$
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