已知 $\tan \theta=2$, 则 $\frac{\sin \left(\frac{\pi}{2}+\theta\right)-\cos (\pi-\theta)}{\sin \left(\frac{\pi}{2}+\theta\right)-\sin (\pi-\theta)}$ 等于 ( )
解析 : $\frac{\sin \left(\frac{\pi}{2}+\theta\right)-\cos (\pi-\theta)}{\sin \left(\frac{\pi}{2}+\theta\right)-\sin (\pi-\theta)}=\frac{\cos \theta+\cos \theta}{\cos \theta-\sin \theta}=\frac{2}{1-\tan \theta}=\frac{2}{1-2}=-2$.

答案: B