If the angle between the lines `sqrt(3)y-x+4=0` and `x+y-6=0` is `theta`, then tan `theta` is equal to.

Options

A. `sqrt(3)+1`
B. `sqrt(3)-1`
C. `2+sqrt(3)`
D. `3+sqrt(2)`

Answer

To find the value of \( \tan \theta \) where \( \theta \) is the angle between the lines given by the equations \( \sqrt{3}y - x + 4 = 0 \) and \( x + y - 6 = 0 \), we will follow these steps:

### Step 1: Convert the equations to slope-intercept form
The slope-intercept form of a line is given by \( y = mx + b \), where \( m \) is the slope.

1. For the first line \( \sqrt{3}y - x + 4 = 0 \):
\[ \sqrt{3}y = x - 4 \implies y = \frac{1}{\sqrt{3}}x - \frac{4}{\sqrt{3}} \]
Thus, the slope \( m_1 = \frac{1}{\sqrt{3}} \).

2. For the second line \( x + y - 6 = 0 \):
\[ y = -x + 6 \]
Thus, the slope \( m_2 = -1 \).

### Step 2: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the slopes into the formula
Substituting \( m_1 = \frac{1}{\sqrt{3}} \) and \( m_2 = -1 \):
\[ \tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - (-1)}{1 + \frac{1}{\sqrt{3}} \cdot (-1)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}} \right| \]

### Step 4: Simplify the expression
1. The numerator:
\[ \frac{1}{\sqrt{3}} + 1 = \frac{1 + \sqrt{3}}{\sqrt{3}} \]

2. The denominator:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

Putting it all together:
\[ \tan \theta = \left| \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \right| = \left| \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \right| \]

### Step 5: Rationalize the denominator
To rationalize \( \frac{1 + \sqrt{3}}{\sqrt{3} - 1} \), multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ \tan \theta = \frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{3 - 1} = \frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{2} \]

### Step 6: Expand the numerator
Expanding \( (1 + \sqrt{3})(\sqrt{3} + 1) \):
\[ = 1 \cdot \sqrt{3} + 1 \cdot 1 + \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1 = \sqrt{3} + 1 + 3 + \sqrt{3} = 4 + 2\sqrt{3} \]

Thus:
\[ \tan \theta = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

### Final Answer
Therefore, \( \tan \theta = 2 + \sqrt{3} \).

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