Use the next trigonometric identities:
[tex]\begin{gathered} \sec (90º-\theta)=\frac{1}{\cos (90º-\theta)} \\ \\ \cos (90º-\theta)=\sin \theta \end{gathered}[/tex]
Then, the sec(90º - θ) is:
[tex]\sec (90º-\theta)=\frac{1}{\sin \theta}[/tex]
The sin(θ) is:
[tex]\sin \theta=\frac{opposite}{hypotenuse}=\frac{5}{13}[/tex]
Then:
[tex]\sec (90º-\theta)=\frac{1}{\frac{5}{13}}=\frac{13}{5}[/tex]
Then, the sec(90º - θ) is 13/5
