Answer
$1$
Work Step by Step
I know that $\cos$ has a period of $2\pi$, hence $\cos{\theta}=\cos{(\theta-2\pi)}.$ Thus $\cos{\frac{37\pi}{18}}=\cos{(\frac{37\pi}{18}-2\pi)}=\cos{\frac{\pi}{18}}$.
I know that $\sec$ is an even function, which means $f(-\theta)=f(\theta).$ Therefore $\sec{-\frac{\pi}{18}}=\sec{\frac{\pi}{18}}$.
Thus: $\cos{\frac{37\pi}{18}}\sec{\frac{-\pi}{18}}=\cos{\frac{\pi}{18}}\sec{\frac{\pi}{18}}=1$.
Because $\cos{\theta}\sec{\theta}=1$.
