We are given that the terminal side of angle $\theta$ intersects the unit circle at the point $\left( -\frac{40}{41}, \frac{9}{41} \right)$.

To find $\sec(\theta)$, recall that:

$\sec(\theta) = \frac{1}{\cos(\theta)}$

Here, the x-coordinate of the point on the unit circle represents $\cos(\theta)$. In this case:

$\cos(\theta) = -\frac{40}{41}$

Thus,

$\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{40}{41}} = -\frac{41}{40}$

Therefore, the simplified and rationalized form of $\sec(\theta)$ is:

$\boxed{-\frac{41}{40}}$

Would you like any further details or have questions on this?

Here are some related questions:

1. What is $\sin(\theta)$ for this point on the unit circle?
2. How would you find $\tan(\theta)$ using the coordinates?
3. Can you explain how the secant function is related to the cosine function?
4. What does the unit circle tell us about the values of trigonometric functions?
5. How would the value of $\sec(\theta)$ change if the point were in another quadrant?

Tip: Remember, secant and cosine are reciprocals, so their signs always follow the same quadrant rules!