If `p=tan^2x+cot^2x,` then which one of the following is correct?

Options

A. `p le 2`
B. `p ge 2`
C. `p lt 2`
D. `p gt 2`

Answer

To solve the problem, we need to analyze the expression given for \( p \):

### Step 1: Write down the expression for \( p \)
We start with the expression:
\[ p = \tan^2 x + \cot^2 x \]

### Step 2: Use the identity for cotangent
Recall that \( \cot x = \frac{1}{\tan x} \). Therefore, we can rewrite \( p \) as:
\[ p = \tan^2 x + \frac{1}{\tan^2 x} \]

### Step 3: Let \( y = \tan^2 x \)
Let \( y = \tan^2 x \). Then we can express \( p \) in terms of \( y \):
\[ p = y + \frac{1}{y} \]

### Step 4: Analyze the function \( f(y) = y + \frac{1}{y} \)
To find the minimum value of \( p \), we consider the function:
\[ f(y) = y + \frac{1}{y} \]
where \( y > 0 \) (since \( \tan^2 x \) is always positive).

### Step 5: Find the derivative and set it to zero
To find the critical points, we take the derivative of \( f(y) \):
\[ f'(y) = 1 - \frac{1}{y^2} \]
Setting the derivative equal to zero gives:
\[ 1 - \frac{1}{y^2} = 0 \Rightarrow y^2 = 1 \Rightarrow y = 1 \]

### Step 6: Evaluate \( f(y) \) at the critical point
Now, we evaluate \( f(y) \) at \( y = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

### Step 7: Determine the behavior of \( f(y) \)
Next, we check the second derivative to confirm it is a minimum:
\[ f''(y) = \frac{2}{y^3} \]
Since \( f''(y) > 0 \) for \( y > 0 \), \( f(y) \) has a minimum at \( y = 1 \).

### Step 8: Conclusion about \( p \)
Thus, the minimum value of \( p \) is:
\[ p \geq 2 \]
This means that \( p \) is always greater than or equal to 2.

### Final Answer
The correct option is:
\[ \text{p is greater than or equal to 2.} \]