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 by rewriting the expression in a more manageable form:
\[ 9\tan^2\theta + 4\cot^2\theta = (3\tan\theta)^2 + (2\cot\theta)^2 \]

### Step 2: Introduce a substitution
Let \(x = \tan\theta\). Then, we have:
\[ \cot\theta = \frac{1}{\tan\theta} = \frac{1}{x} \]
Substituting this into our expression gives:
\[ 9x^2 + 4\left(\frac{1}{x}\right)^2 = 9x^2 + \frac{4}{x^2} \]

### Step 3: Find the derivative
To find the minimum value, we can take the derivative of the function \(f(x) = 9x^2 + \frac{4}{x^2}\) and set it to zero:
\[ f'(x) = 18x - \frac{8}{x^3} \]
Setting the derivative equal to zero gives:
\[ 18x - \frac{8}{x^3} = 0 \]

### Step 4: Solve for \(x\)
Multiplying through by \(x^3\) to eliminate the fraction:
\[ 18x^4 - 8 = 0 \]
\[ 18x^4 = 8 \]
\[ x^4 = \frac{8}{18} = \frac{4}{9} \]
Taking the fourth root gives:
\[ x = \left(\frac{4}{9}\right)^{1/4} = \frac{2^{1/2}}{3^{1/2}} = \frac{\sqrt{2}}{3^{1/2}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \]

### Step 5: Substitute back to find the minimum value
Now we substitute \(x = \frac{\sqrt{6}}{3}\) back into the original function:
\[ f\left(\frac{\sqrt{6}}{3}\right) = 9\left(\frac{\sqrt{6}}{3}\right)^2 + 4\left(\frac{3}{\sqrt{6}}\right)^2 \]
Calculating each term:
\[ = 9 \cdot \frac{6}{9} + 4 \cdot \frac{9}{6} = 6 + 6 = 12 \]

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