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The expression given is:

\[
\frac{\tan \beta}{\cot \beta}
\]

We can simplify this expression as follows:

1. **Rewrite** \(\cot \beta\) in terms of \(\tan \beta\):
   \[
   \cot \beta = \frac{1}{\tan \beta}
   \]

2. **Substitute** this into the expression:
   \[
   \frac{\tan \beta}{\cot \beta} = \frac{\tan \beta}{\frac{1}{\tan \beta}} = \tan \beta \cdot \tan \beta = \tan^2 \beta
   \]

So, the simplified form of \(\frac{\tan \beta}{\cot \beta}\) is \(\tan^2 \beta\).

### Answer:
The correct answer is:
\[
\text{C) } \tan^2 \beta
\]

Would you like further clarification on the steps? 

Here are some related questions to practice:

1. How do you simplify \(\frac{\cot \alpha}{\tan \alpha}\)?
2. What is the relationship between \(\sin^2 \theta + \cos^2 \theta\) and 1?
3. How would you express \(\frac{\sec \theta}{\csc \theta}\) as a single trigonometric function?
4. What is the result of \(\tan^2 \theta + 1\) in terms of \(\sec\)?
5. Can you write \(\cot \theta \cdot \tan \theta\) as a single number?

**Tip:** Remember that rewriting trigonometric functions in terms of each other, like \(\tan\) and \(\cot\), can simplify expressions quickly.

Write the following expression as a single trigonometric function or a power of a trigonometric function: \(\frac{\tan \beta}{\cot \beta}\)

Learn how to simplify the trigonometric expression \(\frac{\tan \beta}{\cot \beta}\) as a single function or a power of a function. This problem covers key concepts in trigonometry, including reciprocal identities and simplification techniques. Suitable for high school students in Grades 10-12.

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