Use the expansion of `tan (A-B) ` to find `tan 15 ^(@) ` without the use of tables, leaving your answer in surd form with an integral denominator.

Answer

To find \( \tan 15^\circ \) using the expansion of \( \tan(A - B) \), we can follow these steps:

### Step 1: Identify \( A \) and \( B \)
We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Thus, we have:
- \( A = 45^\circ \)
- \( B = 30^\circ \)

### Step 2: Use the formula for \( \tan(A - B) \)
The formula for \( \tan(A - B) \) is given by:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Substituting \( A \) and \( B \):
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

### Step 3: Substitute known values
We know:
- \( \tan 45^\circ = 1 \)
- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)

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

### Step 4: Simplify the numerator and denominator
The numerator becomes:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
The denominator becomes:
\[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \]

Putting it all together:
\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 5: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} - 1 \):
\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \]

### Step 6: Simplify the expression
The denominator simplifies using the difference of squares:
\[ (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \]
The numerator expands to:
\[ (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \]
Thus, we have:
\[ \tan 15^\circ = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Final Answer
Therefore, the value of \( \tan 15^\circ \) is:
\[ \tan 15^\circ = 2 - \sqrt{3} \]