[tex]\bf tan(\theta )=9\implies tan(\theta )=\cfrac{\stackrel{opposite}{9}}{\stackrel{adjacent}{1}}\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf c=\sqrt{1^2+9^2}\implies c=\sqrt{82} \\\\[-0.35em] ~\dotfill\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{82}}}\implies \stackrel{\textit{rationalizing the denominator}}{cos(\theta )=\cfrac{1}{\sqrt{82}}\cdot \cfrac{\sqrt{82}}{\sqrt{82}}}\implies cos(\theta )=\cfrac{\sqrt{82}}{82}[/tex]
