Answer
$\sqrt3$
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 on page 111 of the textbook)
A negative angle measure means the terminal side of the angle will move clockwise from the positive x-axis.
Thus, from the positive x-axis, moving the terminal side $\frac{17\pi}{3}$ radians clockwise ends at $\frac{\pi}{3}$ in Figure 11.
The angle $\frac{\pi}{3}$ intersects the unit circle at the point $(\frac{1}{2}, \frac{\sqrt3}{2})$.
This point has:
$x= \frac{1}{2}$
$y=\frac{\sqrt3}{2}$
Thus,
$\tan{(-\frac{17\pi}{3}}
\\= \tan{\frac{5\pi}{3}}
\\=\dfrac{y}{x}
\\=\dfrac{\frac{\sqrt3}{2}}{\frac{1}{2}}
\\=\frac{\sqrt3}{2} \cdot \frac{2}{1}
\\=\sqrt3$
