Which one of the following is the value of tan `75^@`.

Options

A. `(2sqrt3)/(sqrt3 -1)`
B. `(sqrt3 -1)/(sqrt3 +1)`
C. `(sqrt3 +1)/(sqrt3 -1)`
D. `(sqrt3)/(sqrt3 +1)`

Answer

To find the value of \( \tan 75^\circ \), we can use the angle addition formula for tangent. The angle \( 75^\circ \) can be expressed as \( 45^\circ + 30^\circ \).

### Step-by-Step Solution:

1. **Express \( 75^\circ \) as \( 45^\circ + 30^\circ \)**:
\[ \tan 75^\circ = \tan(45^\circ + 30^\circ) \]

2. **Apply the tangent addition formula**:
The formula for \( \tan(a + b) \) is:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Here, \( a = 45^\circ \) and \( b = 30^\circ \).

3. **Substitute the values of \( \tan 45^\circ \) and \( \tan 30^\circ \)**:
We know:
\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Therefore:
\[ \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \]

4. **Simplify the expression**:
The numerator becomes:
\[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \]
The denominator becomes:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

5. **Combine the results**:
Now substituting back into the tangent formula:
\[ \tan 75^\circ = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Final Result:
Thus, the value of \( \tan 75^\circ \) is:
\[ \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]