To find the values of the trigonometric functions for an angle in standard position, we can use the following formulas:
\[ \sin(\alpha) = \frac{y}{r}, \cos(\alpha) = \frac{x}{r}, \tan(\alpha) = \frac{y}{x}, \]
\[ \csc(\alpha) = \frac{r}{y}, \sec(\alpha) = \frac{r}{x}, \cot(\alpha) = \frac{x}{y}, \]
where \( x \) and \( y \) are the coordinates of the point and \( r \) is the distance from the origin to the point, which can be found using the Pythagorean theorem: \( r = \sqrt{x^2 + y^2} \).
In this case, the coordinates of the point are \( x = -2 \) and \( y = 6 \). To find the distance \( r \), we calculate:
\[ r = \sqrt{(-2)^2 + 6^2} = \sqrt{40} = 2\sqrt{10}. \]
Now, we can find the values of the trigonometric functions:
\[ \sin(\alpha) = \frac{y}{r} = \frac{6}{2\sqrt{10}} = \frac{3\sqrt{10}}{10}, \]
\[ \cos(\alpha) = \frac{x}{r} = \frac{-2}{2\sqrt{10}} = -\frac{\sqrt{10}}{10}, \]
\[ \tan(\alpha) = \frac{y}{x} = \frac{6}{-2} = -3, \]
\[ \csc(\alpha) = \frac{r}{y} = \frac{2\sqrt{10}}{6} = \frac{\sqrt{10}}{3}, \]
\[ \sec(\alpha) = \frac{r}{x} = \frac{2\sqrt{10}}{-2} = -\sqrt{10}, \]
\[ \cot(\alpha) = \frac{x}{y} = \frac{-2}{6} = -\frac{1}{3}. \]
Therefore, the exact values of the trigonometric functions for the angle with the terminal point (-2,6) are:
\[ \sin(\alpha) = \frac{3\sqrt{10}}{10}, \cos(\alpha) = -\frac{\sqrt{10}}{10}, \tan(\alpha) = -3, \]
\[ \csc(\alpha) = \frac{\sqrt{10}}{3}, \sec(\alpha) = -\sqrt{10}, \cot(\alpha) = -\frac{1}{3}. \]