\(\frac{\cos (90-\theta)+\sin (180-\theta)}{\cos (180-\theta)-\sin (90-\theta)} \text {. }\)
Using compound angle formulae we have
\begin{aligned}
&\frac{\cos 90 \cos \theta+\sin 90 \sin \theta+\sin 180 \cos \theta-\sin \theta \cos 180}{\cos 180 \cos \theta+\sin 180 \sin \theta-(\sin 90 \cos \theta-\sin \theta \cos \theta)} \\
&=-\tan \theta
\end{aligned}