well, first off let's notice something about angle "t", is between π and 3π/2, or namely the III Quadrant, where both "x" or adjacent side as well as "y" or opposite side are negative.
something else noteworthy is that, the hypotenuse is just a radius length and thus is never negative.
[tex]tan(t)=\cfrac{2}{3}\implies tan(t)=\cfrac{\stackrel{opposite}{-2}}{\underset{adjacent}{-3}}\qquad \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c =\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-3)^2+(-2)^2}\implies c = \sqrt{9+4}\implies c = \sqrt{13}[/tex]
[tex]\rule{34em}{0.25pt}\\\\ sec(t)=\cfrac{\stackrel{hypotenuse}{\sqrt{13}}}{\underset{adjacent}{-3}}~\hfill cos(t)=\cfrac{\stackrel{adjacent}{-3}}{\underset{hypotenuse}{\sqrt{13}}}\implies \cfrac{-3\cdot \sqrt{13}}{\sqrt{13}\cdot \sqrt{13}}\implies \cfrac{-3\sqrt{3}}{13}[/tex]
[tex]sin(t)=\cfrac{\stackrel{opposite}{-2}}{\underset{hypotenuse}{\sqrt{13}}}\implies \cfrac{-2\cdot \sqrt{13}}{\sqrt{13}\cdot \sqrt{13}}\implies \cfrac{-2\sqrt{13}}{13} \\\\\\ csc(t)=\cfrac{\stackrel{hypotenuse}{\sqrt{13}}}{\underset{opposite}{-2}}~\hfill cot=\cfrac{\stackrel{adjacent}{-3}}{\underset{opposite}{-2}}\implies \cfrac{3}{2}[/tex]
