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

If $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A P$, then
  1. $\tan y=\tan x+\tan z$
  2. $\tan y=\tan x-\tan z$
  3. $2 \tan y=\tan x+\tan z$
  4. $2 \tan y=\tan x-\tan z$

Solution

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

Asked in: MHT CET 2021 (24 Sep Shift 1)

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