Expression `tan^(2)alpha + cot^(2)alpha` is:

Options

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

Answer

To solve the expression \( \tan^2 \alpha + \cot^2 \alpha \), we can follow these steps:

### Step 1: Rewrite the expression
We know that:
\[ \cot \alpha = \frac{1}{\tan \alpha} \]
Thus, we can rewrite \( \cot^2 \alpha \) as:
\[ \cot^2 \alpha = \frac{1}{\tan^2 \alpha} \]
Now, substituting this into our expression gives:
\[ \tan^2 \alpha + \cot^2 \alpha = \tan^2 \alpha + \frac{1}{\tan^2 \alpha} \]

### Step 2: Let \( x = \tan^2 \alpha \)
Now, we can simplify our expression further by letting:
\[ x = \tan^2 \alpha \]
So the expression becomes:
\[ x + \frac{1}{x} \]

### Step 3: Analyze the expression \( x + \frac{1}{x} \)
The function \( f(x) = x + \frac{1}{x} \) is defined for \( x > 0 \) (since \( \tan^2 \alpha \) is always non-negative). To find the minimum value of this expression, we can use calculus or the AM-GM inequality.

Using the AM-GM inequality:
\[ x + \frac{1}{x} \geq 2 \]
This inequality holds true for all \( x > 0 \), with equality when \( x = 1 \) (which occurs when \( \tan^2 \alpha = 1 \) or \( \alpha = 45^\circ \)).

### Step 4: Conclusion
Thus, we conclude that:
\[ \tan^2 \alpha + \cot^2 \alpha \geq 2 \]
We can also check specific angles:
- For \( \alpha = 45^\circ \):
\[ \tan^2 45^\circ + \cot^2 45^\circ = 1 + 1 = 2 \]
- For \( \alpha = 30^\circ \):
\[ \tan^2 30^\circ + \cot^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{3})^2 = \frac{1}{3} + 3 = \frac{10}{3} \approx 3.33 \]
- For \( \alpha = 60^\circ \):
\[ \tan^2 60^\circ + \cot^2 60^\circ = (\sqrt{3})^2 + \left(\frac{1}{\sqrt{3}}\right)^2 = 3 + \frac{1}{3} = \frac{10}{3} \approx 3.33 \]

From our analysis, we see that:
\[ \tan^2 \alpha + \cot^2 \alpha \geq 2 \]
and it can take values greater than 2.

### Final Result
Thus, the expression \( \tan^2 \alpha + \cot^2 \alpha \) is always greater than or equal to 2.

---