Since, $\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$
$$
\begin{aligned}
&\Rightarrow \quad(\cos \alpha+\cos \beta)^2-(\sin \alpha+\sin \beta)^2=0 \\
&\Rightarrow \cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta-\sin ^2 \alpha-\sin ^2 \beta \\
&\quad-2 \sin \alpha \sin \beta=0 \\
&\Rightarrow \cos ^2 \alpha-\sin ^2 \alpha+\cos ^2 \beta-\sin ^2 \beta=2(\sin \alpha \sin \beta \\
&\Rightarrow \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta) \quad \text { Hence proved. }
\end{aligned}
$$