The expression `tan^2alpha+cot^2alpha` is

Options

A. `ge2`
B. `le2`
C. `ge-2`
D. none of these

Answer

To solve the expression \( \tan^2 \alpha + \cot^2 \alpha \), we will use the properties of trigonometric functions and the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Recall the definitions**:
- We know that \( \cot \alpha = \frac{1}{\tan \alpha} \).
- Therefore, we can express \( \cot^2 \alpha \) as \( \cot^2 \alpha = \frac{1}{\tan^2 \alpha} \).

2. **Rewrite the expression**:
- The expression can be rewritten as:
\[ \tan^2 \alpha + \cot^2 \alpha = \tan^2 \alpha + \frac{1}{\tan^2 \alpha} \]

3. **Apply the AM-GM inequality**:
- According to the AM-GM inequality, for any two positive numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
- In our case, let \( a = \tan^2 \alpha \) and \( b = \cot^2 \alpha \). Thus:
\[ \frac{\tan^2 \alpha + \cot^2 \alpha}{2} \geq \sqrt{\tan^2 \alpha \cdot \cot^2 \alpha} \]

4. **Simplify the right-hand side**:
- We know that \( \tan \alpha \cdot \cot \alpha = 1 \), hence:
\[ \tan^2 \alpha \cdot \cot^2 \alpha = 1^2 = 1 \]
- Therefore:
\[ \sqrt{\tan^2 \alpha \cdot \cot^2 \alpha} = \sqrt{1} = 1 \]

5. **Combine the results**:
- Substituting back into the AM-GM inequality gives:
\[ \frac{\tan^2 \alpha + \cot^2 \alpha}{2} \geq 1 \]
- Multiplying both sides by 2:
\[ \tan^2 \alpha + \cot^2 \alpha \geq 2 \]

6. **Conclusion**:
- Thus, we conclude that:
\[ \tan^2 \alpha + \cot^2 \alpha \geq 2 \]

### Final Answer:
The expression \( \tan^2 \alpha + \cot^2 \alpha \) is greater than or equal to 2.