What is the value of tan `105^(@)`?

Options

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

Explanation

To find the value of \( \tan 105^\circ \), we can use the angle addition formula for tangent. The angle \( 105^\circ \) can be expressed as \( 45^\circ + 60^\circ \).

### Step-by-Step Solution:

1. **Identify the angles**:
\[ 105^\circ = 45^\circ + 60^\circ \]

2. **Use the tangent addition formula**:
The formula for the tangent of the sum of two angles is:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Here, \( A = 45^\circ \) and \( B = 60^\circ \).

3. **Find the tangent values**:
- \( \tan 45^\circ = 1 \)
- \( \tan 60^\circ = \sqrt{3} \)

4. **Substitute the values into the formula**:
\[ \tan 105^\circ = \tan(45^\circ + 60^\circ) = \frac{\tan 45^\circ + \tan 60^\circ}{1 - \tan 45^\circ \tan 60^\circ} \]
\[ = \frac{1 + \sqrt{3}}{1 - (1)(\sqrt{3})} \]

5. **Simplify the expression**:
\[ = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \]

6. **Rationalize the denominator**:
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]
\[ = \frac{(1 + \sqrt{3})^2}{1 - 3} = \frac{1 + 2\sqrt{3} + 3}{-2} = \frac{4 + 2\sqrt{3}}{-2} \]
\[ = -2 - \sqrt{3} \]

Thus, the value of \( \tan 105^\circ \) is:
\[ \tan 105^\circ = -2 - \sqrt{3} \]