已知 $\tan \theta={ }^{\sqrt{3}}$, 则 $\sin \theta \cos \theta-\cos ^{2} \theta=$
$\sin \vartheta \cos \vartheta-\cos ^{2} \vartheta=\frac{\sin \theta \cos \theta-\cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta}=\frac{\tan \theta-1}{\tan ^{2} \theta+1}=\frac{\sqrt{3}-1}{3+1}=\frac{\sqrt{3}-1}{4}$.