To find the parametric equations for the tangent line at the given point on the curve $X = \cos t$, $y = \sin t$, $z = t$, we'll follow these steps:

1. Find the curve in vector form: The given curve can be represented as: $\mathbf{r}(t) = (\cos t, \sin t, t)$

2. Compute the derivative of the curve: The derivative $\mathbf{r}'(t)$ gives the tangent vector to the curve at any point $t$. Differentiate each component of $\mathbf{r}(t)$: $\mathbf{r}'(t) = (-\sin t, \cos t, 1)$

3. Evaluate the tangent vector at the specific point: The point given corresponds to $t = \frac{5\pi}{6}$. Plugging this value into $\mathbf{r}'(t)$, we get: $\mathbf{r}'\left(\frac{5\pi}{6}\right) = \left(-\sin\left(\frac{5\pi}{6}\right), \cos\left(\frac{5\pi}{6}\right), 1\right)$

   Using trigonometric identities: $\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ So, $\mathbf{r}'\left(\frac{5\pi}{6}\right) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}, 1\right)$

4. Find the coordinates of the point: The coordinates of the point on the curve corresponding to $t = \frac{5\pi}{6}$ are: $\left(\cos\left(\frac{5\pi}{6}\right), \sin\left(\frac{5\pi}{6}\right), \frac{5\pi}{6}\right) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}, \frac{5\pi}{6}\right)$

5. Write the parametric equations of the tangent line: The tangent line at the point is given by: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{r}'(t)$ where $\mathbf{r}_0$ is the point $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}, \frac{5\pi}{6}\right)$ and $\mathbf{r}'(t)$ is the tangent vector $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}, 1\right)$. Therefore, the parametric equations of the tangent line are: $x(t) = -\frac{\sqrt{3}}{2} - \frac{1}{2}t$ $y(t) = \frac{1}{2} - \frac{\sqrt{3}}{2}t$ $z(t) = \frac{5\pi}{6} + t$

These are the parametric equations of the tangent line at the given point.

Would you like further details on any step or have additional questions?

Here are some related questions for practice:

1. How would you find the normal vector at the same point on the curve?
2. What is the equation of the tangent line in vector form?
3. How do you determine the curvature of the curve at the given point?
4. Can you find the angle between the tangent line and a given vector?
5. What would the parametric equations be if the point was at $t = \pi/4$?

Tip: When dealing with trigonometric functions, always double-check the values using reference angles to avoid sign mistakes!