What is `tan 15^(@)` equal to ?

Options

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

Explanation

To find the value of \( \tan 15^\circ \), we can use the tangent subtraction formula. The formula states that:

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

### Step 1: Identify \( A \) and \( B \)

We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Thus, we set:
- \( A = 45^\circ \)
- \( B = 30^\circ \)

### Step 2: Calculate \( \tan A \) and \( \tan B \)

We know:
- \( \tan 45^\circ = 1 \)
- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)

### Step 3: Substitute into the formula

Now, substituting \( A \) and \( B \) into the tangent subtraction formula:

\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

Substituting the values we calculated:

\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]

### Step 4: Simplify the numerator and denominator

**Numerator:**

\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

**Denominator:**

\[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \]

### Step 5: Combine the results

Now, substituting back into the equation gives:

\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \]

This simplifies to:

\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 6: Rationalize the denominator

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Final Answer

Thus, the value of \( \tan 15^\circ \) is:

\[ \tan 15^\circ = 2 - \sqrt{3} \]

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