## Algebra 2 (1st Edition)

$\sqrt 2$
Here, we have $\theta=-\dfrac{15 \pi}{4}$ Here, $\theta'=-\dfrac{15 \pi}{4}+4 \pi=\dfrac{\pi}{4}$ and $\theta''=\theta'=\dfrac{\pi}{4}$ The function $\sec$ is Positive in Quadrant $I$. Now, $\sec \theta=\sec \theta'= \sec \theta''=\sec \dfrac{\pi}{4}=\sqrt 2$