`tan 15^(@)=`

Options

A. `1/3`
B. `sqrt3-2`
C. `2- sqrt3`
D. `2+ sqrt3`

Answer

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

\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]

### Step-by-step solution:

1. **Identify \( a \) and \( b \)**:
We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Here, \( a = 45^\circ \) and \( b = 30^\circ \).

2. **Find \( \tan a \) and \( \tan b \)**:
- \( \tan 45^\circ = 1 \)
- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)

3. **Apply 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 found:
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]

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

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

6. **Combine the results**:
\[ \tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \]
The \( \sqrt{3} \) in the numerator and denominator cancels out:
\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

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

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