First solve for the hypotenuse of the given right triangle using the Pythagorean Theorem
[tex]\begin{gathered} c^2=a^2+b^2 \\ c^2=6^2+8^2 \\ c^2=36+64 \\ c^2=100 \\ \sqrt[]{c^2}=\sqrt[]{100} \\ c=10 \end{gathered}[/tex]The hypotenuse is 10, we can now solve for the six trigonometric ratios.
Solving for the first three trigonometric ratios.
Recall the following ratios
[tex]\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}},\cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta=\frac{\text{opposite}}{\text{adjacent}}[/tex]Relative to angle Θ, we have the following sides
opposite = 8, adjacent = 6, hypotenuse = 10
Therefore, the ratio are the following
[tex]\begin{gathered} \sin \theta=\frac{8}{10},\cos \theta=\frac{6}{10},\tan \theta=\frac{8}{6} \\ \sin \theta=\frac{4}{5},\cos \theta=\frac{3}{5},\tan \theta=\frac{4}{3}\text{ (simplified)} \end{gathered}[/tex]Solving for the reciprocal of the trigonometric ratios
Recall their reciprocal counterparts
[tex]\csc \theta=\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta=\frac{\text{hypotenuse}}{\text{adjacent}},\cot \theta=\frac{\text{adjacent}}{\text{opposite}}[/tex]Substitute, and we have the following
[tex]\begin{gathered} \csc \theta=\frac{10}{8},\sec \theta=\frac{10}{6},\cot \theta=\frac{6}{8} \\ \csc \theta=\frac{5}{4},\sec \theta=\frac{5}{3},\cot \theta=\frac{3}{4}\text{ (simplified)} \end{gathered}[/tex]