in the II Quadrant, let's recall that the adjacent side or cosine is negative whilst the opposite side or sine is positive, thus
[tex]tan(\theta )=-\sqrt{\cfrac{19}{17}}\implies tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{19}}}{\underset{adjacent}{-\sqrt{17}}}\impliedby \qquad \textit{let's find the \underline{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}[/tex]
[tex]c=\sqrt{(-\sqrt{17})^2~~ + ~~(\sqrt{19})^2}\implies c=\sqrt{17+19}\implies c=\sqrt{36}\implies c=6 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{17}}}{\underset{hypotenuse}{6}}~\hfill[/tex]