Find the value of `tan75^(@)`:

Options

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

Answer

To find the value of \( \tan 75^\circ \), we can use the angle addition formula for tangent. We can express \( 75^\circ \) as \( 45^\circ + 30^\circ \).

### Step 1: Use the angle addition formula for tangent
The formula for \( \tan(a + b) \) is given by:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Here, let \( a = 45^\circ \) and \( b = 30^\circ \).

### Step 2: Substitute the values of \( \tan 45^\circ \) and \( \tan 30^\circ \)
We know:
\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Now substituting these values into the formula:
\[ \tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \]

### Step 3: 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 4: Substitute back into the formula
Now substituting back into the formula:
\[ \tan 75^\circ = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \]
This simplifies to:
\[ \tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

### Step 5: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ \tan 75^\circ = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{3 - 1} \]
This gives:
\[ \tan 75^\circ = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

### Final Result
Thus, the value of \( \tan 75^\circ \) is:
\[ \tan 75^\circ = 2 + \sqrt{3} \]