Three positive acute angles `alpha, beta and gamma` satisfy the relation `tan. (beta)/(2)=(1)/(3)cot.(alpha)/(2)and cot.(gamma)/(2)=(1)/(2)(3tan.(alpha)/(2)+cot.(alpha)/(2))`. Then, the value of `alpha+beta+gamma` is equal to

Options

A. `pi`
B. `2pi`
C. `(pi)/(2)`
D. `(3pi)/(2)`

Answer

To solve the problem, we start with the given equations involving the angles α, β, and γ:

1. **Given Equations**:
- \( \tan\left(\frac{\beta}{2}\right) = \frac{1}{3} \cot\left(\frac{\alpha}{2}\right) \)
- \( \cot\left(\frac{\gamma}{2}\right) = \frac{1}{2}\left(3\tan\left(\frac{\alpha}{2}\right) + \cot\left(\frac{\alpha}{2}\right)\right) \)

2. **Convert Cotangent to Tangent**:
- Recall that \( \cot\left(\theta\right) = \frac{1}{\tan\left(\theta\right)} \).
- Thus, we can rewrite the first equation:
\[ \tan\left(\frac{\beta}{2}\right) = \frac{1}{3} \cdot \frac{1}{\tan\left(\frac{\alpha}{2}\right)} = \frac{1}{3\tan\left(\frac{\alpha}{2}\right)} \]

3. **Cross Multiply**:
- Cross multiplying gives:
\[ \tan\left(\frac{\beta}{2}\right) \tan\left(\frac{\alpha}{2}\right) = \frac{1}{3} \]

4. **Using the Tangent Addition Formula**:
- We can express \( \tan\left(\frac{\alpha + \beta}{2}\right) \) using the formula:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{\beta}{2}\right)}{1 - \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)} \]

5. **Substituting Values**:
- Substitute \( \tan\left(\frac{\beta}{2}\right) \) into the formula:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) + \frac{1}{3\tan\left(\frac{\alpha}{2}\right)}}{1 - \frac{1}{3}} = \frac{\tan\left(\frac{\alpha}{2}\right) + \frac{1}{3\tan\left(\frac{\alpha}{2}\right)}}{\frac{2}{3}} \]

6. **Simplifying**:
- Simplifying gives:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{3\left(\tan^2\left(\frac{\alpha}{2}\right) + \frac{1}{3}\right)}{2\tan\left(\frac{\alpha}{2}\right)} \]

7. **Using the Second Equation**:
- The second equation can also be simplified using the same tangent identities.
- Substitute \( \cot\left(\frac{\alpha}{2}\right) \) and \( \tan\left(\frac{\alpha}{2}\right) \) into the second equation:
\[ \cot\left(\frac{\gamma}{2}\right) = \frac{1}{2}\left(3\tan\left(\frac{\alpha}{2}\right) + \frac{1}{\tan\left(\frac{\alpha}{2}\right)}\right) \]

8. **Combining Results**:
- From both equations, we can derive that:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \cot\left(\frac{\gamma}{2}\right) \]
- This implies:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) \tan\left(\frac{\gamma}{2}\right) = 1 \]

9. **Final Relation**:
- The equation \( \tan\left(\frac{\alpha + \beta}{2}\right) \tan\left(\frac{\gamma}{2}\right) = 1 \) leads us to conclude that:
\[ \frac{\alpha + \beta}{2} + \frac{\gamma}{2} = \frac{\pi}{2} \]
- Therefore:
\[ \alpha + \beta + \gamma = \pi \]

10. **Conclusion**:
- The value of \( \alpha + \beta + \gamma \) is \( \pi \).