Given
$\cos A+\cos B+\cos C=0=\sin A+\sin B+\sin C$ ...(i)
Now $\cos A+\cos B+\cos C=0$
$$
\begin{aligned}
& \Rightarrow 2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)+\cos C=0 \\
& \Rightarrow 2 \cos \left(\frac{A-B}{2}\right) \cdot \cos \left(\frac{A+B}{2}\right)=-\cos C
\end{aligned}
$$
$\Rightarrow \quad \cos \left(\frac{A-B}{2}\right)=\frac{-\cos C}{2 \cos \left(\frac{A+B}{2}\right)}$ ...(ii)
Again now $\sin A+\sin B+\sin C=0$ (given)
$$
\Rightarrow \quad 2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)+\sin C=0
$$
$\Rightarrow \quad \cos \left(\frac{A-B}{2}\right)=-\frac{\sin C}{2 \sin \left(\frac{A+B}{2}\right)}$ ...(iii)
From (ii) and (iii), we have
$$
\begin{aligned}
& \Rightarrow \frac{-\cos C}{2 \cos \left(\frac{A+B}{2}\right)}=-\frac{\sin C}{2 \sin \left(\frac{A+B}{2}\right)} \\
& \Rightarrow \tan C=\tan \left(\frac{A+B}{2}\right) \\
& \Rightarrow \mathrm{C}=\frac{A+B}{2}
\end{aligned}
$$
putting $\mathrm{C}=\frac{A+B}{2}$ in equation (ii), we get
$$
\Rightarrow \cos \left(\frac{A-B}{2}\right)=\frac{-\cos \left(\frac{A+B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right)}
$$
$$
\Rightarrow \cos \left(\frac{A-B}{2}\right)=-\frac{1}{2}
$$
we know that
$$
\begin{aligned}
& \cos (A-B)=2 \cdot \cos ^2\left(\frac{A-B}{2}\right)-1 \\
& =2 \cdot\left(-\frac{1}{2}\right)^2-1=-\frac{1}{2}
\end{aligned}
$$