If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$   then  $\tan (x-y)=$ 

$\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$

$\cos x+\cos y=-\cos z, \sin x+\sin y=-\sin z$

Squaring and adding, We get

${\left(\cos x+\cos y\right)}^2+ {\left(\sin x+\sin y\right)}^2=1$

$\left({\cos }^{2}x+{\sin }^{2}x\right)+\left({\cos }^{2}y+{\sin }^{2}y\right)+2\left(\cos x\cos y+\sin x\sin y\right)=1$

$2+2\cos \left(x-y\right)=1$
 $\begin{array}{l}\cos \left(\frac{x-y}{2}\right)=\frac{1}{2},\text{\hspace{0.17em}⇒x-y=120°}\\ \Rightarrow \tan (x-y)=-\sqrt{3}\end{array}$