In a triangle ABC, tanA and tanB satisfy the inequality `sqrt(3)x^(2)-4x+sqrt(3) lt 0`, then which of the following must be correct ?
(where symbols used have usual meanings)

Options

A. `a^(2)+b^(2)-ab lt c^(2)`
B. `a^(2)+b^(2) gt c^(2)`
C. `a^(2)+b^(2)+ab gt c^(2)`
D. `a^(2)+b^(2) lt c^(2)`

Answer

To solve the problem, we need to analyze the given inequality involving the tangent of angles A and B in triangle ABC. The inequality is:

\[ \sqrt{3}x^2 - 4x + \sqrt{3} < 0 \]

### Step 1: Find the roots of the quadratic equation

We start by finding the roots of the quadratic equation:

\[ \sqrt{3}x^2 - 4x + \sqrt{3} = 0 \]

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = \sqrt{3} \), \( b = -4 \), and \( c = \sqrt{3} \).

Calculating the discriminant:

\[ b^2 - 4ac = (-4)^2 - 4(\sqrt{3})(\sqrt{3}) = 16 - 12 = 4 \]

Now substituting into the quadratic formula:

\[ x = \frac{4 \pm \sqrt{4}}{2\sqrt{3}} = \frac{4 \pm 2}{2\sqrt{3}} \]

This gives us two roots:

\[ x_1 = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
\[ x_2 = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 2: Analyze the inequality

The roots of the quadratic are \( x_1 = \sqrt{3} \) and \( x_2 = \frac{1}{\sqrt{3}} \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), so the inequality \( \sqrt{3}x^2 - 4x + \sqrt{3} < 0 \) holds between the roots:

\[ \frac{1}{\sqrt{3}} < x < \sqrt{3} \]

### Step 3: Relate to angles A and B

Since \( \tan A \) and \( \tan B \) must satisfy this inequality, we have:

\[ \frac{1}{\sqrt{3}} < \tan A, \tan B < \sqrt{3} \]

### Step 4: Determine the angle ranges

Taking the inverse tangent:

\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \quad \text{and} \quad \tan^{-1}(\sqrt{3}) = 60^\circ \]

Thus, we have:

\[ 30^\circ < A, B < 60^\circ \]

### Step 5: Find angle C

In triangle ABC, we know:

\[ A + B + C = 180^\circ \]

This implies:

\[ C = 180^\circ - (A + B) \]

Since \( A + B \) is between \( 60^\circ \) and \( 120^\circ \):

\[ 60^\circ < A + B < 120^\circ \]

Thus:

\[ 60^\circ < C < 120^\circ \]

### Step 6: Apply the Cosine Rule

Using the cosine rule:

\[ \cos C = \frac{A^2 + B^2 - C^2}{2AB} \]

Given \( C \) is between \( 60^\circ \) and \( 120^\circ \), we know:

- \( \cos 60^\circ = \frac{1}{2} \) (maximum)
- \( \cos 120^\circ = -\frac{1}{2} \) (minimum)

Thus, we conclude:

1. \( A^2 + B^2 - C^2 < AB \) (for \( C < 120^\circ \))
2. \( A^2 + B^2 + AB > C^2 \) (for \( C > 60^\circ \))

### Conclusion

From the analysis above, we find that the inequalities derived from the angles A, B, and C lead us to conclude that:

- The first option \( A^2 + B^2 - C^2 < AB \) is correct.
- The second option \( A^2 + B^2 + AB > C^2 \) is also correct.