Given point: (3,7)
Opposite side: O
Adjacent side: A
Hypotenuse: H
[tex]\begin{gathered} O=7 \\ A=3 \\ \end{gathered}[/tex]Use pythagorean theorem to find H:
[tex]\begin{gathered} H=\sqrt[]{A^2+O^2} \\ H=\sqrt[]{3^2+7^2} \\ H=\sqrt[]{9+49} \\ H=\sqrt[]{58} \end{gathered}[/tex]Trigonometric functions:
[tex]\begin{gathered} \sin \theta=\frac{O}{H}=\frac{7}{\sqrt[]{58}} \\ \\ \text{Razionalizing the denominator;} \\ \sin \theta=\frac{7}{\sqrt[]{58}}\cdot\frac{\sqrt[]{58}}{\sqrt[]{58}}=\frac{7\sqrt[]{58}}{58} \end{gathered}[/tex][tex]\begin{gathered} \cos \theta=\frac{A}{H}=\frac{3}{\sqrt[]{58}}_{} \\ \\ \text{Razionalizing the denominator:} \\ \cos \theta=\frac{3}{\sqrt[]{58}}_{}\cdot\frac{\sqrt[]{58}}{\sqrt[]{58}}=\frac{3\sqrt[]{58}}{58} \end{gathered}[/tex][tex]\tan \theta=\frac{O}{A}=\frac{7}{3}[/tex][tex]\begin{gathered} \\ \csc \theta=\frac{H}{O}=\frac{\sqrt[]{58}}{7} \end{gathered}[/tex][tex]\sec \theta=\frac{H}{A}=\frac{\sqrt[]{58}}{3}[/tex][tex]\cot \theta=\frac{A}{O}=\frac{3}{7}[/tex]
