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: \(\boxed{\sin(\theta) = \frac{4}{5}, \cos(\theta) = \frac{-3}{5}, \tan(\theta) = \frac{-4}{3}, \csc(\theta) = \frac{5}{4}, \sec(\theta) = \frac{-5}{3}, \cot(\theta) = \frac{-3}{4}}\)