1. **Problem:** Given a point on the terminal side of an angle $\theta$, find the six trigonometric functions: $\sin\theta$, $\cos\theta$, $\tan\theta$, $\csc\theta$, $\sec\theta$, and $\cot\theta$. 2. **Formula and rules:** For a point $(x,y)$ on the terminal side of $\theta$, the radius $r = \sqrt{x^2 + y^2}$. Then: $$\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x}$$ $$\csc\theta = \frac{r}{y}, \quad \sec\theta = \frac{r}{x}, \quad \cot\theta = \frac{x}{y}$$ Note: $r > 0$, and division by zero is undefined. 3. **Given point:** $(-1, 5)$ Calculate $r$: $$r = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$$ 4. **Calculate trig functions:** $$\sin\theta = \frac{5}{\sqrt{26}}$$ $$\cos\theta = \frac{-1}{\sqrt{26}}$$ $$\tan\theta = \frac{5}{-1} = -5$$ $$\csc\theta = \frac{\sqrt{26}}{5}$$ $$\sec\theta = \frac{\sqrt{26}}{-1} = -\sqrt{26}$$ $$\cot\theta = \frac{-1}{5} = -\frac{1}{5}$$ 5. **Summary:** $$\sin\theta = \frac{5}{\sqrt{26}}, \quad \cos\theta = -\frac{1}{\sqrt{26}}, \quad \tan\theta = -5$$ $$\csc\theta = \frac{\sqrt{26}}{5}, \quad \sec\theta = -\sqrt{26}, \quad \cot\theta = -\frac{1}{5}$$ This completes the solution for the first point.