Given that
[tex]\tan \theta=\frac{5}{12}\text{ and 0}^0<\theta<90^0,[/tex]
we are asked to find the value of sec θ. This can be seen below.
Explanation
[tex]\tan \theta=\frac{5}{12}=\frac{\text{opposite}}{Adjacent}[/tex]
We can represent the information in the diagram below;
Therefore we know that
[tex]\sec \theta=\frac{Hypotenuse}{Adjacent}[/tex]
This implies we will need to get the hypotenuse using the Pythagoras theorem. Hence;
[tex]\begin{gathered} (\text{Hypotenuse)}^2=(\text{opposite)}^2+(\text{adjacent)}^2 \\ x^2=12^2+5^2 \\ x=\sqrt[]{144+25} \\ x=\sqrt[]{169} \\ x=13 \end{gathered}[/tex]
Therefore,
[tex]\sec \theta=\frac{\text{Hypotenuse}}{Adjacent}=\frac{x}{12}=\frac{13}{12}[/tex]
Answer:
[tex]\sec \theta=\frac{13}{12}[/tex]
