What is the value of Tan `15^@`?

Options

A. `(sqrt3-1)/(2sqrt2)`
B. `(sqrt3+1)/(2sqrt2)`
C. `(sqrt3-1)/(sqrt3+1)`
D. `(sqrt3+1)/(sqrt3-1)`

Answer

To find the value of \( \tan 15^\circ \), we can use the tangent subtraction formula. Here are the steps to solve the problem:

### Step 1: Express \( \tan 15^\circ \) using known angles
We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Therefore, we can write:
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) \]

### Step 2: Apply the tangent subtraction formula
The formula for \( \tan(a - b) \) is given by:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
In our case, \( a = 45^\circ \) and \( b = 30^\circ \). Thus, we have:
\[ \tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]

### Step 3: Substitute the values of \( \tan 45^\circ \) and \( \tan 30^\circ \)
We know that:
\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Substituting these values into the formula gives:
\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \]

### Step 4: Simplify the expression
Now, we simplify the numerator and denominator:
- **Numerator**:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
- **Denominator**:
\[ 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \]

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: Final result
Thus, the value of \( \tan 15^\circ \) is:
\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]