Answer
$3-\displaystyle \frac{2\sqrt{3}}{3}$
Work Step by Step
reference angles:
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
$120^{o}$ is in quadrant II, where tangent is negative,
reference angle: $180^{o}-120^{o}=60^{o}$
Function Values of Special Angles:
$ \tan 120^{o}=-\tan 60^{o}=-\sqrt{3}$
$240^{o}$ is in quadrant III, where cotangent is positive,
reference angle: $240^{o}-180^{o}=60^{o}$
Function Values of Special Angles:
$\displaystyle \cot 240^{o}=\cot 60^{o}=\frac{\sqrt{3}}{3}$
$\displaystyle \tan^{2}120^{o}-2\cot 240^{o}=(-\sqrt{3})^{2}-2(\frac{\sqrt{3}}{3})$
$=3-\displaystyle \frac{2\sqrt{3}}{3}$
