The point $P=(-2,3)$ on the circle $x^{2}+y^{2}=r^{2}$ is also on the terminal side of an angle $\theta$ in standard position. Find $\sin \theta$, $\cos \theta$, tan $\theta$, $\csc \theta$, sec $\theta$, and $\cot \theta$. (Simplify your answer, including ary radicals. Use integers or fractions for any numbers in the expression)
Solution
Step 1 ：Given the point P=(-2,3) lies on the circle $x^{2}+y^{2}=r^{2}$, we can use the coordinates of the point to find the radius $r$ of the circle using the equation of the circle.
Step 2 ：Substitute x = -2 and y = 3 into the equation to get $r^{2} = (-2)^{2} + 3^{2} = 4 + 9 = 13$, so $r = \sqrt{13}$.
Step 3 ：Using the definitions of the trigonometric functions in terms of x, y, and r, we can find the values of $\sin \theta$, $\cos \theta$, $\tan \theta$, $\csc \theta$, $\sec \theta$, and $\cot \theta$.
Step 4 ：We have $\sin \theta = \frac{y}{r} = \frac{3}{\sqrt{13}}$, $\cos \theta = \frac{x}{r} = -\frac{2}{\sqrt{13}}$, $\tan \theta = \frac{y}{x} = -\frac{3}{2}$.
Step 5 ：Also, $\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{13}}{3}$, $\sec \theta = \frac{1}{\cos \theta} = -\frac{\sqrt{13}}{2}$, and $\cot \theta = \frac{1}{\tan \theta} = -\frac{2}{3}$.
Step 6 ：\(\boxed{\text{Final Answer: } \sin \theta = \frac{3}{\sqrt{13}}, \cos \theta = -\frac{2}{\sqrt{13}}, \tan \theta = -\frac{3}{2}, \csc \theta = \frac{\sqrt{13}}{3}, \sec \theta = -\frac{\sqrt{13}}{2}, \cot \theta = -\frac{2}{3}}\)
