Let's evaluate the given expression:

$\frac{-\sec^2(t)}{-8\sin(t)}$ when $t = -\frac{\pi}{4}$.

Step 1: Simplify the expression

The negative signs cancel each other out, so we can rewrite the expression as:

$\frac{\sec^2(t)}{8\sin(t)}$

Step 2: Find $\sec^2\left(-\frac{\pi}{4}\right)$

Recall that $\sec(t) = \frac{1}{\cos(t)}$, so:

$\sec^2\left(-\frac{\pi}{4}\right) = \left(\frac{1}{\cos\left(-\frac{\pi}{4}\right)}\right)^2$

Since $\cos(-t) = \cos(t)$, and $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we have:

$\sec^2\left(-\frac{\pi}{4}\right) = \left(\frac{1}{\frac{\sqrt{2}}{2}}\right)^2 = 2$

Step 3: Find $\sin\left(-\frac{\pi}{4}\right)$

Since $\sin(-t) = -\sin(t)$, and $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, we have:

$\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

Step 4: Substitute values

Now substitute the values for $\sec^2\left(-\frac{\pi}{4}\right) = 2$ and $\sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ into the expression:

$\frac{2}{8 \cdot \left(-\frac{\sqrt{2}}{2}\right)} = \frac{2}{-4\sqrt{2}} = \frac{-1}{2\sqrt{2}}$

Step 5: Rationalize the denominator

To rationalize the denominator, multiply both the numerator and denominator by $\sqrt{2}$:

$\frac{-1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{-\sqrt{2}}{4}$

Final Answer:

$\frac{-\sqrt{2}}{4}$

Let me know if you'd like further details or have questions! Here are some related questions:

1. What is the process for rationalizing denominators?
2. How does the unit circle help in evaluating trigonometric functions at special angles?
3. What are the properties of the secant function?
4. How do we simplify trigonometric expressions involving negatives?
5. Why is rationalizing the denominator important in mathematics?

Tip: Always remember that $\sin(-t) = -\sin(t)$ and $\cos(-t) = \cos(t)$ when dealing with negative angles!