[tex]\textit{we know that }\theta \textit{ contains the point }(\stackrel{x}{3}~~,~~\stackrel{y}{0})\textit{, let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ r^2=a^2+b^2\implies r=\sqrt{a^2+b^2}\implies r=\sqrt{3^2+0^2} \qquad \begin{cases} r=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ r=\sqrt{3^2}\implies r=3[/tex]
[tex]\rule{34em}{0.25pt}\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse} =\cfrac{y}{r}\implies \cfrac{0}{3}\implies 0 \\\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse} =\cfrac{x}{r}\implies \cfrac{3}{3}\implies 1 \\\\\\ tan(\theta)=\cfrac{opposite}{adjacent} =\cfrac{y}{x}\implies \cfrac{0}{3}\implies 0[/tex]
[tex]cot(\theta)=\cfrac{adjacent}{opposite} =\cfrac{x}{y}\implies \cfrac{3}{0}\implies und efined \\\\\\ csc(\theta)=\cfrac{hypotenuse}{opposite} =\cfrac{r}{y}\implies \cfrac{3}{0}\implies und efined \\\\\\ sec(\theta)=\cfrac{hypotenuse}{adjacent} =\cfrac{r}{x}\implies \cfrac{3}{3}\implies 1[/tex]
