In the interval `(0,pi//2)` the fucntion `f(x)= tan^nx+cot^n`x attains

Options

A. the minimum value which is independent of n
B. a minimum vlaue which is a fuction of n
C. the minimum vlaue which is a function of 1
D. non of these

Explanation

To solve the problem of finding the minimum value of the function \( f(x) = \tan^n x + \cot^n x \) in the interval \( (0, \frac{\pi}{2}) \), we can follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ f(x) = \tan^n x + \cot^n x \]
To facilitate finding the minimum, we can rewrite this expression by adding and subtracting 2:
\[ f(x) = \tan^n x + \cot^n x + 2 - 2 \]
This gives us:
\[ f(x) = \left( \tan^{n/2} x - \cot^{n/2} x \right)^2 + 2 \]

### Step 2: Analyze the Square Term
The term \( \left( \tan^{n/2} x - \cot^{n/2} x \right)^2 \) is always non-negative since it is a square. Therefore, the minimum value of \( f(x) \) occurs when this square term is equal to 0:
\[ \tan^{n/2} x - \cot^{n/2} x = 0 \]

### Step 3: Solve for \( x \)
Setting the square term to zero implies:
\[ \tan^{n/2} x = \cot^{n/2} x \]
This condition holds true when:
\[ \tan x = \cot x \]
This occurs when:
\[ x = \frac{\pi}{4} \]

### Step 4: Find the Minimum Value
Substituting \( x = \frac{\pi}{4} \) back into the function:
\[ f\left(\frac{\pi}{4}\right) = \tan^n\left(\frac{\pi}{4}\right) + \cot^n\left(\frac{\pi}{4}\right) = 1^n + 1^n = 1 + 1 = 2 \]

### Conclusion
Thus, the minimum value of the function \( f(x) \) in the interval \( (0, \frac{\pi}{2}) \) is:
\[ \boxed{2} \]
This minimum value is independent of \( n \).