## Precalculus (6th Edition) Blitzer

The exact value of the expression is $\frac{\sqrt{3}}{3}$.
$\tan \left( -\frac{17\pi }{6} \right)$ lies in quadrant III. Since, $\left( -\frac{17\pi }{6} \right)$, add $4\pi$ (two multiple of $2\pi$) to $\left( -\frac{17\pi }{6} \right)$, the positive co-terminal angle less than $2\pi$. \begin{align} & \theta =\frac{-17\pi }{6}+4\pi \\ & =\frac{-17\pi +24\pi }{6} \\ & =\frac{7\pi }{6} \end{align} The figure shows $\theta =\frac{7\pi }{6}$ in standard position. The angle lies in quadrant III. Therefore, the reference angle is \begin{align} & {\theta }'=\frac{7\pi }{6}-\pi \\ & =\frac{7\pi -6\pi }{6} \\ & =\frac{\pi }{6} \end{align} Hence, $\tan \frac{\pi }{6}=\frac{\sqrt{3}}{3}$ In quadrant I all the trigonometric function are positive. In quadrant II sine and cosecant are positive and all other trigonometric functions are negative. In quadrant III tangent and cotangent are positive and all other trigonometric functions are negative. In quadrant IV cosine and secant are positive and all other trigonometric functions are negative. Here, the tangent is positive in quadrant III. Therefore, $\tan \left( -\frac{17\pi }{6} \right)=\tan \frac{\pi }{6}$ Substitute $\frac{\sqrt{3}}{3}$ for $\tan \frac{\pi }{6}$. $\tan \left( -\frac{17\pi }{6} \right)=\frac{\sqrt{3}}{3}$ The exact value of the expression is $\frac{\sqrt{3}}{3}$.