If A > 0, B > 0 and `A+B=pi/6`, then the minimum value of `tan A + tan B` is

Options

A. `2-sqrt3`
B. `(2)/(sqrt3)`
C. `sqrt3-sqrt2`
D. `4-2sqrt3`

Explanation

To find the minimum value of \( \tan A + \tan B \) given that \( A > 0 \), \( B > 0 \), and \( A + B = \frac{\pi}{6} \), we can follow these steps:

### Step 1: Express \( \tan A + \tan B \)
Using the tangent addition formula, we know that:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Since \( A + B = \frac{\pi}{6} \), we have:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 2: Let \( x = \tan A \) and \( y = \tan B \)
We can rewrite the equation as:
\[ \frac{1}{\sqrt{3}} = \frac{x + y}{1 - xy} \]
Cross-multiplying gives:
\[ x + y = \frac{1 - xy}{\sqrt{3}} \]

### Step 3: Rearranging the equation
Rearranging the above equation leads to:
\[ \sqrt{3}(x + y) + \sqrt{3}xy = 1 \]
This can be rewritten as:
\[ \sqrt{3}xy + \sqrt{3}(x + y) - 1 = 0 \]

### Step 4: Use AM-GM Inequality
To find the minimum value of \( x + y \), we can apply the AM-GM inequality:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Thus,
\[ x + y \geq 2\sqrt{xy} \]

### Step 5: Substitute \( xy \) in terms of \( x + y \)
From the earlier equation, we can express \( xy \) in terms of \( x + y \):
Let \( s = x + y \) and \( p = xy \). Then:
\[ \sqrt{3}p + \sqrt{3}s - 1 = 0 \implies p = \frac{1 - \sqrt{3}s}{\sqrt{3}} \]

### Step 6: Substitute \( p \) back into AM-GM
Substituting \( p \) into the AM-GM inequality gives:
\[ s \geq 2\sqrt{\frac{1 - \sqrt{3}s}{\sqrt{3}}} \]
Squaring both sides leads to:
\[ s^2 \geq 4 \cdot \frac{1 - \sqrt{3}s}{\sqrt{3}} \]
This simplifies to:
\[ s^2 \sqrt{3} \geq 4 - 4\sqrt{3}s \]
Rearranging gives a quadratic inequality in \( s \):
\[ s^2 \sqrt{3} + 4\sqrt{3}s - 4 \geq 0 \]

### Step 7: Solve the quadratic equation
The roots of the quadratic equation \( s^2 \sqrt{3} + 4\sqrt{3}s - 4 = 0 \) can be found using the quadratic formula:
\[ s = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 + 4 \cdot \sqrt{3} \cdot 4}}{2\sqrt{3}} \]
Calculating the discriminant and solving for \( s \) gives the minimum value.

### Step 8: Find the minimum value
After solving the quadratic, we find that the minimum value of \( \tan A + \tan B \) occurs when \( A = B = \frac{\pi}{12} \), leading to:
\[ \tan A + \tan B = 2\tan\left(\frac{\pi}{12}\right) = 2\left(\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) = 2\left(\sqrt{3} - 2\right) \]

Thus, the minimum value of \( \tan A + \tan B \) is:
\[ \boxed{2\sqrt{3} - 4} \]