## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$1$
Recall: $\cos{(\theta+360^{\circ})} = \cos{\theta}$ This means that $\cos{400^{\circ}} = \cos{(40^{\circ}+360^{\circ})} = \cos{40^{\circ}}$ Thus, $\cos{400^{\circ}} \cdot \sec{40^{\circ}} = \cos{40^{\circ}}\cdot \sec{40^{\circ}}$ Since $\sec{\theta} = \dfrac{1}{\cos{\theta}}$, then $\sec{40^{\circ}} = \dfrac{1}{\cos{40^{\circ}}}$ Hence, $\cos{40^{\circ}}\cdot \sec{40^{\circ}} = \cos{40^{\circ}}\cdot \dfrac{1}{\cos{40^{\circ}} }= \boxed{1}$