If `sin alpha +cos alpha =(sqrt(3)+1)/(2), 0 lt alpha lt 2pi`, then possible values `"tan"(alpha)/(2)` can take is/are :

Options

A. `2-sqrt(3)`
B. `(1)/(sqrt(3))`
C. 1
D. `sqrt(3)`

Answer

To solve the problem, we need to find the possible values of \(\frac{\tan(\alpha)}{2}\) given that \(\sin \alpha + \cos \alpha = \frac{\sqrt{3}+1}{2}\) and \(0 < \alpha < 2\pi\).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sin \alpha + \cos \alpha = \frac{\sqrt{3}+1}{2} \]

2. **Square both sides:**
\[ (\sin \alpha + \cos \alpha)^2 = \left(\frac{\sqrt{3}+1}{2}\right)^2 \]
Expanding both sides gives:
\[ \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = \frac{3 + 2\sqrt{3} + 1}{4} \]

3. **Use the Pythagorean identity:**
Since \(\sin^2 \alpha + \cos^2 \alpha = 1\), we can substitute:
\[ 1 + 2 \sin \alpha \cos \alpha = \frac{4 + 2\sqrt{3}}{4} \]

4. **Simplify the equation:**
\[ 2 \sin \alpha \cos \alpha = \frac{4 + 2\sqrt{3}}{4} - 1 \]
\[ 2 \sin \alpha \cos \alpha = \frac{4 + 2\sqrt{3} - 4}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]

5. **Express in terms of \(\sin 2\alpha\):**
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = \frac{\sqrt{3}}{2} \]

6. **Find the angles for \(2\alpha\):**
The equation \(\sin 2\alpha = \frac{\sqrt{3}}{2}\) gives us:
\[ 2\alpha = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 2\alpha = \frac{2\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Dividing by 2:
\[ \alpha = \frac{\pi}{6} + k\pi \quad \text{or} \quad \alpha = \frac{\pi}{3} + k\pi \]

7. **Determine values of \(\alpha\) in the range \(0 < \alpha < 2\pi\):**
- From \(\alpha = \frac{\pi}{6} + k\pi\):
- For \(k=0\): \(\alpha = \frac{\pi}{6}\)
- For \(k=1\): \(\alpha = \frac{7\pi}{6}\)
- From \(\alpha = \frac{\pi}{3} + k\pi\):
- For \(k=0\): \(\alpha = \frac{\pi}{3}\)
- For \(k=1\): \(\alpha = \frac{4\pi}{3}\)

8. **Calculate \(\tan \frac{\alpha}{2}\) for each value of \(\alpha\):**
- For \(\alpha = \frac{\pi}{6}\):
\[ \tan \frac{\alpha}{2} = \tan \frac{\pi/6}{2} = \tan \frac{\pi}{12} \]
- For \(\alpha = \frac{7\pi}{6}\):
\[ \tan \frac{\alpha}{2} = \tan \frac{7\pi/6}{2} = \tan \frac{7\pi}{12} \]
- For \(\alpha = \frac{\pi}{3}\):
\[ \tan \frac{\alpha}{2} = \tan \frac{\pi/3}{2} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \]
- For \(\alpha = \frac{4\pi}{3}\):
\[ \tan \frac{\alpha}{2} = \tan \frac{4\pi/3}{2} = \tan \frac{2\pi}{3} = -\sqrt{3} \]

9. **Final values of \(\tan \frac{\alpha}{2}\):**
The possible values of \(\tan \frac{\alpha}{2}\) are:
\[ \tan \frac{\pi}{12}, \tan \frac{7\pi}{12}, \frac{1}{\sqrt{3}}, -\sqrt{3} \]

### Summary:
The possible values of \(\frac{\tan(\alpha)}{2}\) are \(\tan \frac{\pi}{12}, \tan \frac{7\pi}{12}, \frac{1}{\sqrt{3}}, -\sqrt{3}\).