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 find \( \tan \frac{C}{2} \) given \( \tan \frac{A}{2} = \frac{1}{3} \) and \( \tan \frac{B}{2} = \frac{2}{3} \), we can use the following steps:

### Step 1: Use the angle sum property of triangles
In a triangle, the sum of the angles is \( 180^\circ \). Therefore, we have:
\[ A + B + C = 180^\circ \]
This implies:
\[ C = 180^\circ - (A + B) \]

### Step 2: Express \( \frac{A + B}{2} \)
From the above equation, we can express \( \frac{A + B}{2} \):
\[ \frac{A + B}{2} = 90^\circ - \frac{C}{2} \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan\left(\frac{A + B}{2}\right) = \tan\left(90^\circ - \frac{C}{2}\right) = \cot\left(\frac{C}{2} \]
Thus, we have:
\[ \tan\left(\frac{A + B}{2}\right) = \cot\left(\frac{C}{2}\right) \]

### Step 4: Find \( \tan\left(\frac{A + B}{2}\right) \)
Using the tangent addition formula:
\[ \tan\left(\frac{A + B}{2}\right) = \frac{\tan\frac{A}{2} + \tan\frac{B}{2}}{1 - \tan\frac{A}{2} \tan\frac{B}{2}} \]
Substituting the values:
\[ \tan\frac{A}{2} = \frac{1}{3}, \quad \tan\frac{B}{2} = \frac{2}{3} \]
We calculate:
\[ \tan\left(\frac{A + B}{2}\right) = \frac{\frac{1}{3} + \frac{2}{3}}{1 - \left(\frac{1}{3} \cdot \frac{2}{3}\right)} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7} \]

### Step 5: Relate \( \tan\left(\frac{C}{2}\right) \) to \( \tan\left(\frac{A + B}{2}\right) \)
Since \( \tan\left(\frac{A + B}{2}\right) = \cot\left(\frac{C}{2}\right) \), we have:
\[ \cot\left(\frac{C}{2}\right) = \frac{9}{7} \]

### Step 6: Find \( \tan\left(\frac{C}{2}\right) \)
Using the relationship \( \tan\theta = \frac{1}{\cot\theta} \):
\[ \tan\left(\frac{C}{2}\right) = \frac{1}{\cot\left(\frac{C}{2}\right)} = \frac{1}{\frac{9}{7}} = \frac{7}{9} \]

Thus, the value of \( \tan\left(\frac{C}{2}\right) \) is:
\[ \boxed{\frac{7}{9}} \]