ANSWER :
The answer is :
[tex]-\frac{5\sqrt{10}}{9}[/tex]
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 :
[tex]\cot\gamma=\frac{9}{13}=\frac{x}{y}[/tex]
Since x and y are both negatives, x = -9 and y = -13
We can have the triangle :
The hypotenuse will be :
[tex]\begin{gathered} c=\sqrt{(-9)^2+(-13)^2} \\ c=5\sqrt{10} \end{gathered}[/tex]
We are asked to find the value of sec y.
In an angle with a terminal point (x, y)
The secant is :
[tex]\sec\gamma=\frac{\text{ hypotenuse}}{x}[/tex]
The hypotenuse is 5√10 and x = -9
The value of sec will be :
[tex]\begin{gathered} \sec\gamma=\frac{5\sqrt{10}}{-9} \\ \\ =-\frac{5\sqrt{10}}{9} \end{gathered}[/tex]
