ANSWER :
The answer is :
![-(5√(10))/(9)](https://img.qammunity.org/2023/formulas/mathematics/college/p4sokw5mzzaymf8mg8qf05keravszldjy7.png)
EXPLANATION :
Note that cotangent is only positive when the angle is in the first or third quadrant.
Since y is not in the first quadrant, it must be in the third quadrant.
So the x and y are both negative.
An angle with a terminal point (x, y)
The cotangent is x/y
We can equate :
![\cot\gamma=(9)/(13)=(x)/(y)](https://img.qammunity.org/2023/formulas/mathematics/college/k92zhrxpwq1i76hv3nu11jyourjwzjb8nd.png)
Since x and y are both negatives, x = -9 and y = -13
We can have the triangle :
The hypotenuse will be :
![\begin{gathered} c=√((-9)^2+(-13)^2) \\ c=5√(10) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bfdh15yye9x369myb9yzfq0ztum3jbweag.png)
We are asked to find the value of sec y.
In an angle with a terminal point (x, y)
The secant is :
![\sec\gamma=\frac{\text{ hypotenuse}}{x}](https://img.qammunity.org/2023/formulas/mathematics/college/6p45eys0ajbe51nchey3myl9mr43gyfhae.png)
The hypotenuse is 5√10 and x = -9
The value of sec will be :
![\begin{gathered} \sec\gamma=(5√(10))/(-9) \\ \\ =-(5√(10))/(9) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dktw0zn7ovlxf4uin901uiuuggqyw58ke0.png)