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

$1$
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, $\sec{\left(\dfrac{37 \pi}{18} \right)}\cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \dfrac{1}{\cos{\left(\dfrac{37 \pi}{18} \right)}} \cdot \cos{\left(\dfrac{37 \pi}{18} \right)} = \boxed{1}$