Given the angle, θ is an angle in quadrant II
[tex]\cos \theta=-\frac{8}{9}[/tex]
From the given equation:
[tex]\cos \theta=\frac{adjacent}{hypotenuse}[/tex]
So, the adjacent = 8 and the hypotenuse = 9
Using the Pythagorean to find the opposite side
[tex]\text{opposite}=\sqrt[]{9^2-8^2}=\sqrt[]{17}[/tex]
we will find csc θ and cot θ
so,
[tex]\begin{gathered} \csc \theta=\frac{hypotenuse}{opposite}=\frac{9}{\sqrt[]{17}}=\frac{9\sqrt[]{17}}{17} \\ \\ \cot \theta=\frac{adjacent}{opposite}=-\frac{8}{\sqrt[]{17}}=-\frac{8\sqrt[]{17}}{17} \end{gathered}[/tex]
