We have to find θ, where:
[tex]\begin{gathered} \sec (\theta)=\frac{1}{\cos (\theta)}=-\frac{13}{5} \\ \tan (\theta)<0 \end{gathered}[/tex]
We can transform the first equation as:
[tex]\sec (\theta)=-\frac{13}{5}\Rightarrow\cos (\theta)=\frac{1}{-\frac{13}{5}}=-\frac{5}{13}[/tex]
As the tangent of the angle is negative, and the tangent is:
[tex]\tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)}[/tex]
the sine of the angle has to be positive (a quotient between a positive and a negative number is negative).
Then, if the sine is positive and the cosine is negative, then theta is in the second quadrant, between 90° and 180°.
We then can calculate the angle as:
[tex]\cos (\theta)=-\frac{5}{13}\Rightarrow\theta=\arccos (-\frac{5}{13})\approx113\degree[/tex]
Answer: θ is approximately 113°, located in the second quadrant.
