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Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​

$$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$$
$$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$$