use the (x,y) coordinates in the figure to find the value of tan 11π/6 or state that the expression is…

Explanation:

Step1: Rewrite the angle

We know that $\frac{11\pi}{6}=2\pi-\frac{\pi}{6}$. The tangent - function has a period of $\pi$, so $\tan(\frac{11\pi}{6})=\tan(2\pi - \frac{\pi}{6})$. Since $\tan(x + k\pi)=\tan(x)$ for any real - number $x$ and integer $k$, and $\tan(2\pi - \alpha)=-\tan\alpha$, then $\tan(\frac{11\pi}{6})=-\tan\frac{\pi}{6}$.

Step2: Recall the value of $\tan\frac{\pi}{6}$

The tangent of an angle $\theta$ in the unit - circle is defined as $\tan\theta=\frac{y}{x}$, where $(x,y)$ is the point on the unit - circle corresponding to the angle $\theta$. For $\theta = \frac{\pi}{6}$, the coordinates on the unit - circle are $(\frac{\sqrt{3}}{2},\frac{1}{2})$, so $\tan\frac{\pi}{6}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Step3: Calculate $\tan\frac{11\pi}{6}$

Since $\tan(\frac{11\pi}{6})=-\tan\frac{\pi}{6}$, then $\tan\frac{11\pi}{6}=-\frac{\sqrt{3}}{3}$.

Answer:

A. $\tan\frac{11\pi}{6}=-\frac{\sqrt{3}}{3}$