To simplify $\tan^2(\theta) \cdot \cot^2(\theta)$ to a single trigonometric function, we'll use the reciprocal identities for tangent and cotangent:

Step-by-step Solution:

1. Recall the reciprocal identities: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \quad \text{and} \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ Therefore: $\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)} \quad \text{and} \quad \cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)}$

2. Multiply the two expressions: $\tan^2(\theta) \cdot \cot^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)} \cdot \frac{\cos^2(\theta)}{\sin^2(\theta)}$

3. Simplify: $\frac{\sin^2(\theta)}{\cos^2(\theta)} \cdot \frac{\cos^2(\theta)}{\sin^2(\theta)} = 1$

Thus, the simplified expression is: $\tan^2(\theta) \cdot \cot^2(\theta) = 1$

Final Answer:

$1$

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Related Questions:

1. What is the reciprocal identity for $\sec(\theta)$?
2. How can you express $\csc(\theta)$ in terms of sine?
3. Prove that $\tan(\theta) \cdot \cot(\theta) = 1$.
4. Simplify $\sec^2(\theta) - \tan^2(\theta)$ to a single trigonometric function.
5. How does the Pythagorean identity relate to reciprocal identities?

Tip:

Always start by rewriting trigonometric functions in terms of sine and cosine when simplifying; it often reveals the relationships clearly.