Find the value of `tan15^@`.

Answer

To find the value of \( \tan 15^\circ \), we can use the tangent subtraction formula. The formula for \( \tan(A - B) \) is given by:

\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

### Step-by-Step Solution:

1. **Identify A and B**:
We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Thus, let \( A = 45^\circ \) and \( B = 30^\circ \).

2. **Apply the Tangent Subtraction Formula**:
Using the formula, we have:
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

3. **Substitute Known Values**:
We know that:
- \( \tan 45^\circ = 1 \)
- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)

Substituting these values into the formula gives:
\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]

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

5. **Simplify the Denominator**:
The denominator becomes:
\[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \]

6. **Combine the Results**:
Now substituting back, we have:
\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \]
This simplifies to:
\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

7. **Rationalize the Expression**:
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} - 1 \):
\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} \]

8. **Final Simplification**:
This simplifies to:
\[ \tan 15^\circ = 2 - \sqrt{3} \]

### Final Answer:
\[ \tan 15^\circ = 2 - \sqrt{3} \]