## Precalculus (10th Edition)

$1$
I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Hence: $\cos{400^\circ}\sec{40^\circ}=\cos{(400^\circ-360^\circ)}\sec{40^\circ}=\cos{40^\circ}\sec{40^\circ}$. We also know that $\cos{\theta}\sec{\theta}=1$. Thus: $\cos{40^\circ}\sec{40^\circ}=1$.