Answer
$\color{blue}{\sec{(\frac{23\pi}{6})}=\dfrac{2\sqrt3}{3}}$
Work Step by Step
Note that $\dfrac{23\pi}{6}$ is coterminal with:
$\dfrac{23\pi}{6} - 2\pi=\dfrac{23\pi}{6} - \dfrac{12\pi}{6}=\dfrac{11\pi}{6}$
Figure 13 on page 579 of this book shows that the unit circle point $(\frac{\sqrt3}{2}, -\frac{1}{2})$ corresponds to the real number $\dfrac{23\pi}{6}$ or $\dfrac{11\pi}{6}$.
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:
$\color{blue}{\sec{(\frac{23\pi}{6})}=\sec{\frac{11\pi}{6}}=\dfrac{1}{\frac{\sqrt3}{2}}=1 \cdot \frac{2}{\sqrt3}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{(\sqrt3)(\sqrt3)}=\frac{2\sqrt3}{3}}$
