Answer:
To answer this question, we will use the following diagram as reference:
Now, recall that in a right triangle the trigonometric functions are defined as follows:

Also, recall that:

From the diagram, we get that:

substituting r=√x²+y² we get:
![\sin \theta=\frac{y}{\sqrt[]{x^2+y^2}}.](https://img.qammunity.org/2023/formulas/mathematics/college/xhj4jjjpyng6z2t9aggvloyfgnldowkpvl.png)
Analogously, we get that:
![\begin{gathered} \cos \theta=\frac{x}{\sqrt[]{x^2+y^2}}, \\ \text{tan}\theta=(y)/(x), \\ \csc \theta=\frac{\sqrt[]{x^2+y^2}}{y}, \\ \sec \theta=\frac{\sqrt[]{x^2+y^2}}{x}, \\ \cot \theta=(x)/(y)\text{.} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6jo0qs3fo8fcpu5nl2hvaypp4w1m0j5nfw.png)