Find the minimum value of `9tan^2theta+4cot^2theta`

Explanation

To find the minimum value of the expression \( 9\tan^2\theta + 4\cot^2\theta \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = 9\tan^2\theta + 4\cot^2\theta \]

### Step 2: Substitute \( \cot^2\theta \)
Using the identity \( \cot\theta = \frac{1}{\tan\theta} \), we can express \( \cot^2\theta \) in terms of \( \tan^2\theta \):
\[ \cot^2\theta = \frac{1}{\tan^2\theta} \]
Thus, we can rewrite \( y \) as:
\[ y = 9\tan^2\theta + 4\left(\frac{1}{\tan^2\theta}\right) \]

### Step 3: Let \( x = \tan^2\theta \)
Let \( x = \tan^2\theta \). Then, we can rewrite the expression as:
\[ y = 9x + \frac{4}{x} \]

### Step 4: Find the derivative
To find the minimum value, we need to take the derivative of \( y \) with respect to \( x \) and set it to zero:
\[ \frac{dy}{dx} = 9 - \frac{4}{x^2} \]
Setting the derivative equal to zero:
\[ 9 - \frac{4}{x^2} = 0 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ \frac{4}{x^2} = 9 \]
\[ 4 = 9x^2 \]
\[ x^2 = \frac{4}{9} \]
\[ x = \frac{2}{3} \quad (\text{since } x = \tan^2\theta \text{ must be positive}) \]

### Step 6: Substitute back to find \( y \)
Now substitute \( x = \frac{2}{3} \) back into the expression for \( y \):
\[ y = 9\left(\frac{2}{3}\right) + \frac{4}{\frac{2}{3}} = 6 + 6 = 12 \]

### Step 7: Verify it's a minimum
To confirm that this is a minimum, we can check the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{8}{x^3} \]
Since \( x > 0 \), \( \frac{d^2y}{dx^2} > 0 \), indicating that \( y \) has a minimum at this point.

### Conclusion
Thus, the minimum value of \( 9\tan^2\theta + 4\cot^2\theta \) is:
\[ \boxed{12} \]