Answer
$\frac{2\sqrt3}{3}$
Work Step by Step
RECALL:
$\sin{s} = y
\\\cos{s} = x
\\\tan{s} = \frac{y}{x}
\\\cot{s} = \frac{x}{y}
\\\sec{s} = \frac{1}{x}
\\\csc{s}=\frac{1}{y}$
(refer to Figure 11 , page 111 of the textbook)
Note that $\frac{23\pi}{6} = 3\frac{5}{6}\pi$. This is $\frac{\pi}{6}$ radians short of $4\pi$, which is 2 full rotations from the positive x-axis.
The terminal side of this angle is the same as the terminal side of $\frac{11\pi}{6}$.
This angle intersects the unit circle at the point $(\frac{\sqrt3}{2}, -\frac{1}{2})$.
This point has:
$x= \frac{\sqrt3}{2}$
$y=-\frac{1}{2}$
Thus,
$\sec{\frac{23\pi}{6}}
\\= \sec{\frac{11\pi}{6}}
\\=\frac{1}{x}
\\=\frac{1}{\frac{\sqrt3}{2}}
\\= 1 \cdot \frac{2}{\sqrt3}
\\=\frac{2}{\sqrt3}$
Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain:
$=\frac{2}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3}
\\=\frac{2\sqrt3}{3}$
