Find the value of `tan75^(@)`

Answer

To find the value of \( \tan 75^\circ \), we can use the angle addition formula for tangent, which states:

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

### Step 1: Identify \( A \) and \( B \)
We can express \( 75^\circ \) as \( 45^\circ + 30^\circ \). Thus, we set:
- \( A = 45^\circ \)
- \( B = 30^\circ \)

### Step 2: Apply the Tangent Addition Formula
Using the formula:

\[ \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \]

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

Substituting these values into the formula gives:

\[ \tan(75^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \]

### Step 4: Simplify the Numerator and Denominator
The numerator becomes:

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

The denominator becomes:

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

### Step 5: Combine the Results
Now we can write:

\[ \tan(75^\circ) = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \]

The \( \sqrt{3} \) in the numerator and denominator cancels out:

\[ \tan(75^\circ) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Final Answer
Thus, the value of \( \tan 75^\circ \) is:

\[ \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]