From the given right-angle triangle, we are provided with the following sides;
[tex]\begin{gathered} Hypotenuse=17 \\ Opposite=8 \end{gathered}[/tex]
We will make use of the Pythagoras theorem to obtain the third side, which is the Adjacent side.
Thus, we have:
[tex]\begin{gathered} H^2=O^2+A^2 \\ 17^2=8^2+A^2 \\ 289=64+A^2 \\ 289-64=A^2 \\ 225=A^2 \\ A=\sqrt[]{225} \\ A=15 \end{gathered}[/tex]
i)
[tex]\begin{gathered} Sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}} \\ Sin\theta=\frac{8}{17} \end{gathered}[/tex]
ii)
[tex]\begin{gathered} Cos\theta=\frac{Adjacent}{\text{Hypotenuse}} \\ \text{Cos}\theta=\frac{15}{17} \end{gathered}[/tex]
iii)
[tex]\begin{gathered} \text{Tan}\theta=\frac{\text{Opposite}}{\text{Adjacent}} \\ \text{Tan}\theta=\frac{8}{15} \end{gathered}[/tex]
iv)
[tex]\begin{gathered} co\sec (\csc )\text{ is the inverse/reciprocal of sine} \\ \text{csc }\theta=\frac{Hypotenuse}{\text{Opposite}} \\ \csc \theta=\frac{17}{8} \\ \end{gathered}[/tex]
v)
[tex]\begin{gathered} \sec \theta\text{ is the inverse/reciprocal of cos }\theta \\ \sec \theta=\frac{\text{Hypotenuse}}{\text{Adjacent}} \\ \sec \theta=\frac{17}{15} \end{gathered}[/tex]
vi)
[tex]\begin{gathered} \cot \theta\text{ is the inverse/reciprocal of tan }\theta \\ \cot \theta=\frac{\text{Adjacent }}{\text{Opposite}} \\ \cot \theta=\frac{15}{8} \end{gathered}[/tex]
