已知 $\tan \alpha=2$, 则 $\frac{\sin (\alpha-\pi)+\cos (\pi-\alpha)}{\sin \left(\frac{\pi}{2}-\alpha\right)+\cos \left(\frac{\pi}{2}+\alpha\right)}$
$\frac{\sin (\alpha-\pi)+\cos (\pi-\alpha)}{\sin \left(\frac{\pi}{2}-\alpha\right)+\cos \left(\frac{\pi}{2}+\alpha\right)}=\frac{-\sin \alpha-\cos \alpha}{\cos \alpha-\sin \alpha}=\frac{-\tan \alpha-1}{1-\tan \alpha}=\frac{-2-1}{1-2}=3$

故选 A.