# Let theta be an angle in quadrant II such that cos theta= -8/9.

Let theta be an angle in quadrant II such that $\mathrm{cos}\theta =\frac{-8}{9}.$
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Karla Bautista
As we know that
$\mathrm{cos}\theta =\frac{-8}{9}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\theta =\sqrt{1-\left(-\frac{8}{9}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1-\frac{64}{81}}=\frac{\sqrt{17}}{9}\phantom{\rule{0ex}{0ex}}\mathrm{csc}\left(\theta \right)=\frac{1}{\mathrm{sin}\theta }=\frac{1}{\frac{1}{\frac{\sqrt{17}}{9}}}\phantom{\rule{0ex}{0ex}}=\frac{9}{17}\sqrt{17}$
$\mathrm{cot}\theta =\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=-\frac{8}{17}\sqrt{17}$