We are given the expression:

$(\sin 20^\circ) \cdot (\cot 20^\circ) \cdot (\sec^2 0^\circ)$

We need to simplify it such that the final answer contains only one trigonometric function and no fractions. Let's proceed step by step.

Step 1: Rewrite $\cot 20^\circ$

Recall that $\cot \theta = \frac{\cos \theta}{\sin \theta}$. Applying this, we get:

$\cot 20^\circ = \frac{\cos 20^\circ}{\sin 20^\circ}$

Substituting this into the original expression:

$(\sin 20^\circ) \cdot \frac{\cos 20^\circ}{\sin 20^\circ} \cdot \sec^2 0^\circ$

Step 2: Simplify the expression

Now, notice that $\sin 20^\circ$ in the numerator and denominator cancel each other:

$\cos 20^\circ \cdot \sec^2 0^\circ$

Step 3: Simplify $\sec^2 0^\circ$

Recall that $\sec 0^\circ = \frac{1}{\cos 0^\circ}$, and since $\cos 0^\circ = 1$, we have:

$\sec^2 0^\circ = 1^2 = 1$

Substituting this into the expression:

$\cos 20^\circ \cdot 1 = \cos 20^\circ$

Final Answer:

The simplified expression is:

$\cos 20^\circ$

Thus, the given trigonometric expression simplifies to $\cos 20^\circ$.

Would you like further details or have any questions about the steps? Here are some related questions:

1. How do you derive the identity for $\cot \theta$?
2. What is the geometric meaning of $\sec \theta$?
3. How does $\sec^2 0^\circ$ simplify to 1?
4. Can you explain how the cancellation of $\sin 20^\circ$ works?
5. What happens if you replace 20° with another angle, say 45°?

Tip: Always check for fundamental identities when simplifying trigonometric expressions; they often help reduce complexity quickly.