To find \(\tan\left(\frac{11\pi}{6}\right)\), let's first determine the coordinates of the angle in the unit circle.
- The angle \(\frac{11\pi}{6}\) is in the fourth quadrant.
- The reference angle for \(\frac{11\pi}{6}\) can be found by calculating: \[ 2\pi - \frac{11\pi}{6} = \frac{12\pi}{6} - \frac{11\pi}{6} = \frac{\pi}{6} \]
- The coordinates for \(\frac{\pi}{6}\) are \(\left(\cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right)\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\). Since \(\frac{11\pi}{6}\) is in the fourth quadrant, the cosine is positive and the sine is negative:
- Thus, the coordinates for \(\frac{11\pi}{6}\) are \(\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\).
Now, the tangent is calculated as follows: \[ \tan\left(\frac{11\pi}{6}\right) = \frac{y}{x} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}. \]
Therefore, the correct answer corresponds to:
B. Tan(11pi/6) = y/x = -1/2/sqrt3/2 = -sqrt3/3.