Answer
$$\sin \theta = -\frac{2\sqrt{13}}{13}, \quad \cos \theta = \frac{3\sqrt{13}}{13}, \\ \tan \theta =-\frac{2}{3}, \quad \cot \theta =- \frac{3}{2}, \\ \sec \theta = \frac{\sqrt{13}}{3}, \quad \csc \theta = - \frac{\sqrt{13}}{2}$$
Work Step by Step
According to the information given in the question, for the point $P$ we have$$x=3, \quad y= -2, \quad r=\sqrt{x^2+y^2}=\sqrt{3^2+(-2)^2}=\sqrt{13}.$$ So we can find the value of the trigonometric functions as follows.$$\sin \theta =\frac{y}{r}=\frac{-2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}, \quad \cos \theta =\frac{x}{r}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}, \\ \tan \theta =\frac{y}{x}=\frac{-2}{3}=-\frac{2}{3}, \quad \cot \theta =\frac{x}{y}=\frac{3}{-2}=-\frac{3}{2}, \\ \sec \theta =\frac{r}{x}=\frac{\sqrt{13}}{3}, \quad \csc \theta =\frac{r}{y}=\frac{\sqrt{13}}{-2}=-\frac{\sqrt{13}}{2}$$
