Answer
$1$
Work Step by Step
Recall:
Secant is an even function so $\because \sec{(-\theta)} = \sec{\theta}$.
This means that
$\sec{\left(-\dfrac{\pi}{18} \right)} = \sec{\left(\dfrac{\pi}{18} \right)}$
and
$\sec{\left(-\dfrac{\pi}{18} \right)} \cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \sec{\left(\dfrac{\pi}{18} \right)}\cdot \cos{\left(\dfrac{37 \pi}{18} \right)}$
Since $\sec{(\theta+2 \pi)} = \sec{\theta}$, then
$\sec{\left(\dfrac{\pi}{18} \right)} = \sec{\left(\dfrac{\pi}{18}+2 \pi \right)} = \sec{\left(\dfrac{37 \pi}{18} \right)}$
Thus,
$\sec{\left(\dfrac{\pi}{18} \right)}\cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \sec{\left(\dfrac{37 \pi}{18} \right)}\cdot \cos{\left(\dfrac{37 \pi}{18} \right)}$
Recall that:
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
Hence,
$\sec{\left(\dfrac{37 \pi}{18} \right)}\cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \dfrac{1}{\cos{\left(\dfrac{37 \pi}{18} \right)}} \cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \boxed{1}$
