Let A, B and C are the angles of a plain triangle and `tan(A/2)=1/3,tan(B/2)=2/3` .then `tan(C/2)` is equal to

Options

A. `7//9`
B. `2//9`
C. `1//3`
D. `2//3`

Answer

To solve the problem, we need to find \( \tan\left(\frac{C}{2}\right) \) given that \( \tan\left(\frac{A}{2}\right) = \frac{1}{3} \) and \( \tan\left(\frac{B}{2}\right) = \frac{2}{3} \).

### Step-by-Step Solution:

1. **Use the Angle Sum Property**:
We know that in a triangle, the sum of the angles is \( A + B + C = 180^\circ \). Therefore, we can express \( C \) in terms of \( A \) and \( B \):
\[ C = 180^\circ - (A + B) \]

2. **Divide by 2**:
Dividing the equation by 2 gives us:
\[ \frac{C}{2} = 90^\circ - \frac{A + B}{2} \]

3. **Use the Tangent Addition Formula**:
We can use the tangent addition formula:
\[ \tan\left(\frac{A + B}{2}\right) = \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right)}{1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right)} \]

4. **Substitute the Known Values**:
Substitute \( \tan\left(\frac{A}{2}\right) = \frac{1}{3} \) and \( \tan\left(\frac{B}{2}\right) = \frac{2}{3} \):
\[ \tan\left(\frac{A + B}{2}\right) = \frac{\frac{1}{3} + \frac{2}{3}}{1 - \left(\frac{1}{3} \cdot \frac{2}{3}\right)} \]

5. **Calculate the Numerator and Denominator**:
The numerator is:
\[ \frac{1}{3} + \frac{2}{3} = 1 \]
The denominator is:
\[ 1 - \left(\frac{1}{3} \cdot \frac{2}{3}\right) = 1 - \frac{2}{9} = \frac{7}{9} \]

6. **Combine the Results**:
Now we can find \( \tan\left(\frac{A + B}{2}\right) \):
\[ \tan\left(\frac{A + B}{2}\right) = \frac{1}{\frac{7}{9}} = \frac{9}{7} \]

7. **Relate to \( \tan\left(\frac{C}{2}\right) \)**:
Since \( \frac{C}{2} = 90^\circ - \frac{A + B}{2} \), we have:
\[ \tan\left(\frac{C}{2}\right) = \cot\left(\frac{A + B}{2}\right) = \frac{1}{\tan\left(\frac{A + B}{2}\right)} = \frac{1}{\frac{9}{7}} = \frac{7}{9} \]

### Final Answer:
Thus, we find that:
\[ \tan\left(\frac{C}{2}\right) = \frac{7}{9} \]