Prove that

$\frac{\cos (\pi+\theta) \cdot \cos (-\theta)}{\cos (\pi-\theta) \cdot \cos \left(\frac{\pi}{2}+\theta\right)}=\frac{-\cos \theta \cdot \cos \theta}{-\cos \theta \cdot-\sin \theta}$

$\Rightarrow \frac{\cos \theta}{-\sin \theta}=-\cot \theta$

$\left(\right.$ Using $\cos (\pi-\theta)=-\cos \theta$ and $\left.\cos \left(\frac{\pi}{2}-\theta\right)=-\sin \theta, \cos (-\theta)=-\cos \theta\right)$

(In III quadrantcosx is negative, $\cos (\pi+\theta)=-\cos \theta)$