SOLUTION
Let us make a simple diagram of the position of the angle using the points (5, -9).
Now, from the right angled triagle in the diagram, let us find the hypotenues, using the Pythagoras theorem.
From the Pythagoras theorem
[tex]\begin{gathered} h^2=5^2+9^2 \\ h^2=25+81 \\ h=\sqrt[]{106} \end{gathered}[/tex]
Now,
[tex]\begin{gathered} \cos \theta=\frac{adjacent}{\text{hypotenuse }}=\frac{5}{\sqrt[]{106}} \\ \cos \theta=\frac{5}{\sqrt[]{106}} \\ \text{Also, in this quadrant, cos}\theta\text{ is positive } \end{gathered}[/tex]
Then
[tex]\begin{gathered} \sec \theta=\frac{1}{\cos \theta} \\ \sec \theta=\frac{\sqrt[]{106}}{5} \end{gathered}[/tex]
Hence, the answer is
[tex]\sec \theta=\frac{\sqrt[]{106}}{5}[/tex]
