If $\theta$ = 120º , then

$$\begin{gathered} \sin 120=\sin \left(180-60°\right)=\sin 60=\frac{\sqrt{3}}{2} \left(\because Sine is positive in II quadrant\right) \\ \cos 120=\cos \left(180-60°\right)=-\cos 60=\frac{-1}{2} \left(\because \cos ine is negative in II quadrant\right) \\ \mathrm{co}t120=\mathrm{co}t \left(180-60°\right)=-\mathrm{co}t 60=\frac{-1}{\sqrt{3}} \left(\because \mathrm{co}t is negative in II quadrant\right) \\ \tan 120=\tan \left(180-60°\right)=-\tan 60=-\sqrt{3} \left(\because \tan is negative in II quadrant\right) \end{gathered}$$