$ ext{Find the value of } an( heta) ext{ for } heta ext{ on the unit circle}$

To find the value of $\tan(\theta)$ for $\theta$ on the unit circle, consider the point $P(x, y)$ on the circle corresponding to the angle $\theta$. The tangent of $\theta$ is given by the ratio of the y-coordinate to the x-coordinate, i.e., $\tan(\theta) = \frac{y}{x}$. For example, if $\theta = \frac{\pi}{4}$, the coordinates of the corresponding point on the unit circle are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. Therefore,

$\tan \left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$

Given $ heta$ on the unit circle, we can use the coordinates of the point $P(x, y)$ to determine $ an( heta)$. If $ heta = frac{5pi}{6}$, the coordinates of $P$ are $left( -frac{sqrt{3}}{2}, frac{1}{2}
ight)$. The tangent value is the quotient of the y-coordinate by the x-coordinate:

$ an left(frac{5pi}{6}
ight) = frac{frac{1}{2}}{-frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3}.$

To determine $ an( heta)$, we use the point $(x, y)$ corresponding to $ heta$ on the unit circle. For instance, if $ heta = frac{3pi}{4}$, the coordinates are $left( -frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight)$. Thus,

$ an left(frac{3pi}{4}
ight) = frac{frac{sqrt{2}}{2}}{-frac{sqrt{2}}{2}} = -1.$

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