To find the coordinates [tex]\((x, y)\)[/tex] on the terminal ray of the angle [tex]\(\theta\)[/tex], we start by using the given trigonometric functions:
[tex]\[\csc \theta = \frac{13}{12}, \quad \sec \theta = -\frac{13}{5}, \quad \cot \theta = -\frac{5}{12}\][/tex]
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- The cosecant function [tex]\(\csc \theta\)[/tex] is the reciprocal of the sine function [tex]\(\sin \theta\)[/tex]:
[tex]\[\sin \theta = \frac{1}{\csc \theta} = \frac{1}{\frac{13}{12}} = \frac{12}{13}\][/tex]
- The secant function [tex]\(\sec \theta\)[/tex] is the reciprocal of the cosine function [tex]\(\cos \theta\)[/tex]:
[tex]\[\cos \theta = \frac{1}{\sec \theta} = \frac{1}{-\frac{13}{5}} = -\frac{5}{13}\][/tex]
2. Verify the value of [tex]\(\cot \theta\)[/tex]:
- The cotangent function [tex]\(\cot \theta\)[/tex] is the ratio of the cosine function to the sine function:
[tex]\[\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-\frac{5}{13}}{\frac{12}{13}} = -\frac{5}{12}\][/tex]
This matches the given [tex]\(\cot \theta = -\frac{5}{12}\)[/tex].
3. Determine the coordinates [tex]\((x, y)\)[/tex]:
- We know that [tex]\(\sin \theta = \frac{y}{r}\)[/tex] and [tex]\(\cos \theta = \frac{x}{r}\)[/tex], where [tex]\(r\)[/tex] (the hypotenuse) is 13 (same as the denominator in [tex]\(\csc \theta\)[/tex]).
- Solving for [tex]\(y\)[/tex]:
[tex]\[
\sin \theta = \frac{y}{r} = \frac{12}{13} \implies y = 12
\][/tex]
- Solving for [tex]\(x\)[/tex]:
[tex]\[
\cos \theta = \frac{x}{r} = -\frac{5}{13} \implies x = -5
\][/tex]
Given the values:
- [tex]\(x = -5\)[/tex]
- [tex]\(y = 12\)[/tex]
Hence, the coordinates of the point [tex]\((x, y)\)[/tex] on the terminal ray of the angle [tex]\(\theta\)[/tex] are [tex]\(\boxed{(-5, 12)}\)[/tex].
