What is `tan(7(1)/(2))^(@)` equal to ?

Options

A. `sqrt(6)+sqrt(3)-sqrt(2)+2`
B. `sqrt(6)+sqrt(3)+sqrt(2)+2`
C. `sqrt(6)-sqrt(3)+sqrt(2)-2`
D. `sqrt(6)+sqrt(3)+sqrt(2)-2`

Explanation

To solve the problem, we need to find the value of \( \tan\left(7 \frac{1}{2}^\circ\right) \).

### Step-by-Step Solution:

1. **Convert the angle to radians (if necessary)**:
Since the angle is given in degrees, we can directly use it in trigonometric functions without conversion.

2. **Use the tangent addition formula**:
We can express \( \tan\left(7 \frac{1}{2}^\circ\right) \) as:
\[ \tan\left(7.5^\circ\right) = \tan\left(\frac{15^\circ}{2}\right) \]

3. **Apply the half-angle formula for tangent**:
The half-angle formula for tangent is:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)} \]
Here, \( \theta = 15^\circ \), so we have:
\[ \tan\left(7.5^\circ\right) = \frac{1 - \cos(15^\circ)}{\sin(15^\circ)} \]

4. **Calculate \( \cos(15^\circ) \) and \( \sin(15^\circ) \)**:
We can use the cosine and sine subtraction formulas:
\[ \cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ \]
\[ = \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

Similarly for \( \sin(15^\circ) \):
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ \]
\[ = \left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

5. **Substitute \( \cos(15^\circ) \) and \( \sin(15^\circ) \) back into the tangent formula**:
Now substitute these values into the tangent formula:
\[ \tan(7.5^\circ) = \frac{1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \]

6. **Simplify the expression**:
The numerator becomes:
\[ 1 - \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{2\sqrt{2} - (\sqrt{3} + 1)}{2\sqrt{2}} = \frac{2\sqrt{2} - \sqrt{3} - 1}{2\sqrt{2}} \]

Thus, we have:
\[ \tan(7.5^\circ) = \frac{2\sqrt{2} - \sqrt{3} - 1}{\sqrt{3} - 1} \]

7. **Rationalize the denominator**:
Multiply numerator and denominator by \( \sqrt{3} + 1 \):
\[ \tan(7.5^\circ) = \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{2} \]

8. **Final simplification**:
After simplifying, we will arrive at the final value of \( \tan(7.5^\circ) \).