[tex]\bf cos^{-1}\left( \cfrac{1}{8} \right)=\theta \qquad \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{8}}\impliedby \textit{let's find the \underline{opposite}} \\\\\\ \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{8^2-1^1}=b\implies \pm\sqrt{63}=b ~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\pm\sqrt{63}}}{\stackrel{adjacent}{1}} \\\\\\ ~\hspace{34em}[/tex]
[tex]\bf \stackrel{\textit{keeping in mind that}}{tan\left(cos^{-1}\left( \frac{1}{8} \right) \right)}\implies tan(\theta )[/tex]
