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The point given below is on the terminal side of an angle $\theta$. Find the exact value of each of the six trigonometric functions of $\theta$.
$(-9,12)$
$$\sin \theta =$$
(Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression. Rationalize all denominators.)

Step 1 ：Given the point (-9,12), we can think of -9 as the adjacent side (x) and 12 as the opposite side (y).

Step 2 ：We can find the hypotenuse (r) using the Pythagorean theorem: $r=\sqrt{x^2+y^2}=\sqrt{(-9{)}^2+{12}^2}=\sqrt{81+144}=\sqrt{225}=15$.

Step 3 ：Now we can find the six trigonometric functions:

Step 4 ：$\sin (\theta )=\frac{y}{r}=\frac{12}{15}=\frac{4}{5}$

Step 5 ：$\cos (\theta )=\frac{x}{r}=\frac{-9}{15}=\frac{-3}{5}$

Step 6 ：$\tan (\theta )=\frac{y}{x}=\frac{12}{-9}=\frac{-4}{3}$

Step 7 ：$\csc (\theta )=\frac{r}{y}=\frac{15}{12}=\frac{5}{4}$

Step 8 ：$\sec (\theta )=\frac{r}{x}=\frac{15}{-9}=\frac{-5}{3}$

Step 9 ：$\cot (\theta )=\frac{x}{y}=\frac{-9}{12}=\frac{-3}{4}$

Step 10 ：So, the exact values of the six trigonometric functions of $\theta$ are $\sin (\theta )=\frac{4}{5}$, $\cos (\theta )=\frac{-3}{5}$, $\tan (\theta )=\frac{-4}{3}$, $\csc (\theta )=\frac{5}{4}$, $\sec (\theta )=\frac{-5}{3}$, and $\cot (\theta )=\frac{-3}{4}$.

Step 11 ：Final Answer: $\boxed{{\displaystyle \sin (\theta )=\frac{4}{5},\cos (\theta )=\frac{-3}{5},\tan (\theta )=\frac{-4}{3},\csc (\theta )=\frac{5}{4},\sec (\theta )=\frac{-5}{3},\cot (\theta )=\frac{-3}{4}}}$