find tan(11π/6). (1 point) o tan(11π/6)=y/x = (-1/2)/(-√3/2)=√3/3 o tan(11π/6)=x/y = (-√3/2)/(-1/2)=√3 o…

Explanation:

Step1: Recall tangent - unit - circle relationship

The formula for $\tan\theta$ in terms of the coordinates $(x,y)$ on the unit - circle is $\tan\theta=\frac{y}{x}$, where $\theta$ is the angle measured counter - clockwise from the positive $x$ - axis.

Step2: Locate the angle $\frac{11\pi}{6}$ on the unit - circle

The angle $\frac{11\pi}{6}$ is equivalent to $330^{\circ}$ ($\frac{11\pi}{6}\times\frac{180^{\circ}}{\pi}=330^{\circ}$). The coordinates of the point on the unit - circle corresponding to the angle $\theta = \frac{11\pi}{6}$ are $(x=\frac{\sqrt{3}}{2},y =-\frac{1}{2})$.

Step3: Calculate the tangent value

Using the formula $\tan\theta=\frac{y}{x}$, substitute $x = \frac{\sqrt{3}}{2}$ and $y=-\frac{1}{2}$ into it. Then $\tan(\frac{11\pi}{6})=\frac{y}{x}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$.

Answer:

$\tan(\frac{11\pi}{6})=\frac{y}{x}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{\sqrt{3}}{3}$, so the correct option is $\tan(\frac{11\pi}{6})=\frac{y}{x}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{\sqrt{3}}{3}$ (the first option in the multiple - choice list).