Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 8

$\sqrt 2$

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, $\sec \theta=\sec \theta'= \sec \theta''=\sec \dfrac{\pi}{4}=\sqrt 2$