In `(0,pi//2)tan^(m)x+cot^(m)x` attains

Options

A. a minimum value which is independent of m
B. a minimum value which is a function of m
C. the minimum value of 2
D. the minimum value at the some point indiependent of m.

Answer

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

### Step 1: Rewrite the Expression
We start with the expression:
\[ y = \tan^m x + \cot^m x \]
Using the identity \( \cot x = \frac{1}{\tan x} \), we can rewrite the expression as:
\[ y = \tan^m x + \left(\frac{1}{\tan x}\right)^m = \tan^m x + \frac{1}{\tan^m x} \]

### Step 2: Let \( t = \tan^m x \)
Now, let \( t = \tan^m x \). Therefore, the expression becomes:
\[ y = t + \frac{1}{t} \]
where \( t > 0 \) since \( x \) is in the interval \( (0, \frac{\pi}{2}) \).

### Step 3: Find the Minimum Value of \( y \)
To find the minimum value of \( y \), we can differentiate it with respect to \( t \):
\[ \frac{dy}{dt} = 1 - \frac{1}{t^2} \]
Setting the derivative equal to zero to find critical points:
\[ 1 - \frac{1}{t^2} = 0 \implies t^2 = 1 \implies t = 1 \]
(Note: \( t = -1 \) is not valid since \( t > 0 \).)

### Step 4: Verify Minimum Value
Now, we need to check if this critical point gives a minimum. We can use the second derivative test:
\[ \frac{d^2y}{dt^2} = \frac{2}{t^3} \]
Since \( t > 0 \), \( \frac{d^2y}{dt^2} > 0 \) indicates that \( y \) has a local minimum at \( t = 1 \).

### Step 5: Calculate the Minimum Value
Substituting \( t = 1 \) back into the expression for \( y \):
\[ y = 1 + \frac{1}{1} = 2 \]

### Conclusion
Thus, the minimum value of \( \tan^m x + \cot^m x \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \) is \( 2 \), and it occurs at \( x = \frac{\pi}{4} \).

### Final Answer
The minimum value of \( \tan^m x + \cot^m x \) is \( 2 \).

---