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
$2 \tan y=\tan x-\tan z$
$\tan y=\tan x+\tan z$
$2 \tan y=\tan x+\tan z$
$\tan y=\tan x-\tan z$
Solution
As $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in A.P.
$\therefore \sin (z+x-y)-\sin (y+z-x)=\sin (x+y-z)-\sin (z+x-y)$
$\Rightarrow 2 \cos z \sin (x-y)=2 \cos x \sin (y-z)$
$\Rightarrow \cos z \sin (x-y)=\cos x \sin (y-z)$
$\Rightarrow \frac{\cos z \sin (x-y)}{\cos x \cos y \cos z}=\frac{\cos x \sin (y-z)}{\cos x \cos y \cos z}$
$\Rightarrow \frac{\sin (x-y)}{\cos x \cos y}=\frac{\sin (y-z)}{\cos y \cos z}$
Using $\frac{\sin (A-B)}{\cos A \cos B}=\tan A-\tan B$
$\Rightarrow \tan x-\tan y=\tan y-\tan z$
$\Rightarrow \tan y-\tan x=\tan z-\tan y$
$\therefore \tan x, \tan y \operatorname{and} \tan z \operatorname{are}$ in A.P.