## Precalculus (10th Edition)

The identity says: $cot(\theta)=\frac{cos\theta}{sin\theta}$ and $csc(\theta)=\frac{1}{sin(\theta)}$. Hence $cos(\theta)\cdot csc(\theta)=cos(\theta)\cdot \frac{1}{sin(\theta)}=\frac{cos(\theta)}{sin\theta}=cot\theta.$