to be an angle in quadrant II such that cos O= -5/8Find the exact values of csc O and cot O.

let's recall that the cosine/adjacent is negative on the II Quadrant, whilst the sine/opposite is positive on that same quadrant, also let's recall that the hypotenuse is never negative, since it's just a radius distance.[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-5}}{\stackrel{hypotenuse}{8}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-(-5)^2}=b\implies \pm\sqrt{64-25}=b\implies \pm\sqrt{39}=b\implies \stackrel{II~Quadrant}{+\sqrt{39}=b} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf csc(\theta )\implies \cfrac{\stackrel{hypotenuse}{8}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{8}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{8\sqrt{39}}{39} \\\\\\ cot(\theta )\implies \cfrac{\stackrel{adjacent}{-5}}{\stackrel{opposite}{\sqrt{39}}}\implies \cfrac{-5}{\sqrt{39}}\cdot \cfrac{\sqrt{39}}{\sqrt{39}}\implies \cfrac{-5\sqrt{39}}{39}[/tex]