The minium value of `4tan^(2)theta+9cot^(2)theta` is equal to

Options

A. 0
B. 5
C. 12
D. 13

Answer

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

### Step 1: Rewrite the expression
We start with the expression:
\[ 4\tan^2\theta + 9\cot^2\theta \]
We can rewrite \(\cot^2\theta\) in terms of \(\tan^2\theta\):
\[ \cot^2\theta = \frac{1}{\tan^2\theta} \]
Let \( x = \tan^2\theta \). Then, the expression becomes:
\[ 4x + \frac{9}{x} \]

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

### Step 3: Solve for x
Rearranging the equation:
\[ \frac{9}{x^2} = 4 \]
Multiplying both sides by \( x^2 \):
\[ 9 = 4x^2 \]
Dividing by 4:
\[ x^2 = \frac{9}{4} \]
Taking the square root:
\[ x = \frac{3}{2} \]

### Step 4: Find the second derivative
To confirm that this point is a minimum, we check the second derivative:
\[ f''(x) = \frac{18}{x^3} \]
Since \( x > 0 \), \( f''(x) > 0 \), indicating a local minimum.

### Step 5: Calculate the minimum value
Now we substitute \( x = \frac{3}{2} \) back into the original expression:
\[ f\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right) + \frac{9}{\frac{3}{2}} \]
Calculating each term:
\[ = 4 \times \frac{3}{2} + 9 \times \frac{2}{3} \]
\[ = 6 + 6 = 12 \]

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