Answer
$\color{blue}{\sec{\frac{5\pi}{4}}=-\sqrt2}$
Work Step by Step
Figure 13 on page 579 of this book shows that the unit circle point $(\frac{-\sqrt2}{2}, -\frac{\sqrt2}{2})$ corresponds to the real number $\dfrac{5\pi}{4}$.
RECALL:
(i) $\cos{s} = x$
(ii) $\sin{s} = y$
(iii) $\tan{s}=\dfrac{y}{x}$
(iv) $\sec{s} =\dfrac{1}{x}$
(v) $\csc{s} = \dfrac{1}{y}$
(vi) $\cot{s} = \dfrac{x}{y}$
Use the coordinates of the unit circle point above and the formula in (iv) above to obtain:
$\color{blue}{\sec{\frac{5\pi}{4}}=\dfrac{1}{\frac{-\sqrt2}{2}}=1 \cdot \frac{-2}{\sqrt2}=-\frac{2}{\sqrt2}=-\frac{2(\sqrt2)}{(\sqrt2)(\sqrt2)}=-\sqrt2}$
