The problem asks us to compute all six trigonometric functions for an angle whose terminal side passes through the point $(\sqrt{6}, -\sqrt{10})$. Here’s the solution:

Step 1: Determine the radius ($r$)

The radius $r$ is the distance from the origin to the point $(x, y)$, given by: $r = \sqrt{x^2 + y^2}$ Substitute $x = \sqrt{6}$ and $y = -\sqrt{10}$: $r = \sqrt{(\sqrt{6})^2 + (-\sqrt{10})^2} = \sqrt{6 + 10} = \sqrt{16} = 4$

Step 2: Define the trigonometric functions

Using $x = \sqrt{6}$, $y = -\sqrt{10}$, and $r = 4$, we compute:

1. Sine ($\sin \theta$): $\sin \theta = \frac{y}{r} = \frac{-\sqrt{10}}{4}$

2. Cosine ($\cos \theta$): $\cos \theta = \frac{x}{r} = \frac{\sqrt{6}}{4}$

3. Tangent ($\tan \theta$): $\tan \theta = \frac{y}{x} = \frac{-\sqrt{10}}{\sqrt{6}} = -\sqrt{\frac{10}{6}} = -\sqrt{\frac{5}{3}}$

4. Cosecant ($\csc \theta$): $\csc \theta = \frac{r}{y} = \frac{4}{-\sqrt{10}} = -\frac{4}{\sqrt{10}} = -\frac{4\sqrt{10}}{10} = -\frac{2\sqrt{10}}{5}$

5. Secant ($\sec \theta$): $\sec \theta = \frac{r}{x} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$

6. Cotangent ($\cot \theta$): $\cot \theta = \frac{x}{y} = \frac{\sqrt{6}}{-\sqrt{10}} = -\sqrt{\frac{6}{10}} = -\sqrt{\frac{3}{5}}$

Final Answers:

1. $\sin \theta = -\frac{\sqrt{10}}{4}$
2. $\cos \theta = \frac{\sqrt{6}}{4}$
3. $\tan \theta = -\sqrt{\frac{5}{3}}$
4. $\csc \theta = -\frac{2\sqrt{10}}{5}$
5. $\sec \theta = \frac{2\sqrt{6}}{3}$
6. $\cot \theta = -\sqrt{\frac{3}{5}}$

Would you like me to expand on any part of this solution or clarify something?

Here are some related questions:

1. How do you determine the quadrant of the angle from the point’s coordinates?
2. What is the significance of the sign of each trigonometric function in different quadrants?
3. How does the unit circle help in visualizing these trigonometric values?
4. Can trigonometric identities verify the relationships among these six functions?
5. How would the values change if the point were in a different quadrant?

Tip: Always reduce radicals when simplifying fractions with square roots for clean and accurate results.