let's recall that on the III Quadrant sine/y is negative and cosine/y is negative, now, the hypotenuse/radius is never negative, since it's just a radius unit.
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{2}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{2^2-(-1)^2}=a\implies \pm\sqrt{4-1}=a\implies \pm\sqrt{3}=a\implies \stackrel{\textit{III Quadrant}}{-\sqrt{3}=a} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{-\sqrt{3}}}\implies \stackrel{\textit{rationalizing the denominator}}{tan(\theta )=\cfrac{-1}{-\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} }\implies tan(\theta )=\cfrac{\sqrt{3}}{3}[/tex]
