Answer
$\displaystyle \sin 120^{\mathrm{o}}=\frac{\sqrt{3}}{2}$
$\displaystyle \cot 120^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$
$\displaystyle \cos 120^{\mathrm{o}}=-\frac{1}{2}$
$\tan 120^{\mathrm{o}}=-\sqrt{3}$
$\sec 120^{\mathrm{o}}=-2$
$\displaystyle \csc 120^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quadrant: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta
\end{array}\right]$
$120^{0}$ is in quadrant II,
so the reference angle is $180^{o}-120^{0}=60^{\mathrm{o}}$.
Since $120^{0}$ is in quadrant II,
the sine and cosecant are positive,
the cosine, tangent, cotangent and secant are negative.
From the Function Values of Special Angles, for $60^{\mathrm{o}}$:
$\displaystyle \sin 120^{\mathrm{o}}=\sin 60^{\mathrm{o}}=\frac{\sqrt{3}}{2}$
$\displaystyle \cot 120^{\mathrm{o}}=-\cot 60^{\mathrm{o}}=-\frac{\sqrt{3}}{3}$
$\displaystyle \cos 120^{\mathrm{o}}=-\cos 60^{\mathrm{o}}=-\frac{1}{2}$
$\tan 120^{\mathrm{o}}=-\tan 60^{\mathrm{o}}=-\sqrt{3}$
$\sec 120^{\mathrm{o}}=-\sec 60^{\mathrm{o}}=-2$
$\displaystyle \csc 120^{\mathrm{o}}=\csc 60^{\mathrm{o}}=\frac{2\sqrt{3}}{3}$
