[tex](\frac{5}{13},y)[/tex]
This is given point in the fourth quadrant.
In this point, the adjacent is
[tex]\frac{5}{13}[/tex]
The opposite is y.
Find hypotenuse h using the pythagorean theorem:
[tex]\begin{gathered} h^2=(\frac{5}{13})^2+y^2 \\ h=\sqrt[]{\frac{25}{169}+y^2} \end{gathered}[/tex][tex]\sec (\theta)[/tex]
is equal to hypotenuse by adjacent.
[tex]\cot (\theta)[/tex]
is equal to adjacent by opposite.
In the fourth quadrant,
[tex]\sec \theta[/tex]
is positive , and
[tex]\cot \theta[/tex]
is negative.
So,
[tex]\begin{gathered} \sec \theta=\frac{h}{\frac{5}{13}} \\ =\frac{13\sqrt[]{\frac{25}{169}+y^2}}{5} \\ \cot \theta=\frac{\frac{5}{13}}{-y} \\ =-\frac{5}{13y} \end{gathered}[/tex]
