If `sin^(2)alpha+sin^(2)beta+cos^(2)gamma=3` then find `tan((alpha+beta+gamma)/(4))+cot((alpha+beta+gamma)/(4))`.

Answer

To solve the problem step by step, we start with the given equation:

1. **Given Equation**:
\[ \sin^2 \alpha + \sin^2 \beta + \cos^2 \gamma = 3 \]

2. **Understanding the Range**:
- We know that the range of \(\sin^2\) and \(\cos^2\) functions is from 0 to 1.
- Therefore, for the sum to equal 3, each term must equal its maximum value of 1:
\[ \sin^2 \alpha = 1, \quad \sin^2 \beta = 1, \quad \cos^2 \gamma = 1 \]

3. **Finding Values of Angles**:
- From \(\sin^2 \alpha = 1\), we conclude:
\[ \sin \alpha = 1 \implies \alpha = 90^\circ \]
- From \(\sin^2 \beta = 1\), we conclude:
\[ \sin \beta = 1 \implies \beta = 90^\circ \]
- From \(\cos^2 \gamma = 1\), we conclude:
\[ \cos \gamma = 1 \implies \gamma = 0^\circ \]

4. **Calculating the Sum of Angles**:
- Now, we compute \(\alpha + \beta + \gamma\):
\[ \alpha + \beta + \gamma = 90^\circ + 90^\circ + 0^\circ = 180^\circ \]

5. **Finding the Required Expression**:
- We need to find:
\[ \tan\left(\frac{\alpha + \beta + \gamma}{4}\right) + \cot\left(\frac{\alpha + \beta + \gamma}{4}\right) \]
- Substituting the sum of angles:
\[ \tan\left(\frac{180^\circ}{4}\right) + \cot\left(\frac{180^\circ}{4}\right) = \tan(45^\circ) + \cot(45^\circ) \]

6. **Calculating \(\tan(45^\circ)\) and \(\cot(45^\circ)\)**:
- We know:
\[ \tan(45^\circ) = 1 \quad \text{and} \quad \cot(45^\circ) = 1 \]
- Therefore:
\[ \tan(45^\circ) + \cot(45^\circ) = 1 + 1 = 2 \]

7. **Final Answer**:
- The final result is:
\[ \boxed{2} \]