In a triangle `ABC, CD` is the bisector of the angle C. If `cos (C/2)` has the value `1/3 and l(CD)=6,` then `(1/a+1/b)` has the value equal to -

Explanation

To solve the problem step by step, we will use the properties of triangles and the angle bisector theorem.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a triangle \( ABC \) with \( CD \) as the angle bisector of angle \( C \). We know that \( \cos(C/2) = \frac{1}{3} \) and the length of \( CD = 6 \). We need to find the value of \( \frac{1}{a} + \frac{1}{b} \).

2. **Using the Area of the Triangle**: The area of triangle \( ABC \) can be expressed as:
\[ \text{Area} = \frac{1}{2} \times a \times b \times \sin C \]
where \( a = BC \) and \( b = AC \).

3. **Finding Areas of Sub-Triangles**: The area of triangle \( ACD \) is:
\[ \text{Area}_{ACD} = \frac{1}{2} \times a \times CD \times \sin\left(\frac{C}{2}\right) = \frac{1}{2} \times a \times 6 \times \sin\left(\frac{C}{2}\right) \]
The area of triangle \( DCB \) is:
\[ \text{Area}_{DCB} = \frac{1}{2} \times b \times CD \times \sin\left(\frac{C}{2}\right) = \frac{1}{2} \times b \times 6 \times \sin\left(\frac{C}{2}\right) \]

4. **Total Area of Triangle \( ABC \)**: Therefore, the total area can be expressed as:
\[ \text{Area}_{ABC} = \text{Area}_{ACD} + \text{Area}_{DCB} = \frac{1}{2} \times a \times 6 \times \sin\left(\frac{C}{2}\right) + \frac{1}{2} \times b \times 6 \times \sin\left(\frac{C}{2}\right) \]

5. **Simplifying the Areas**: Factoring out common terms:
\[ \text{Area}_{ABC} = 3 \sin\left(\frac{C}{2}\right) (a + b) \]

6. **Relating Areas**: We also have:
\[ \text{Area}_{ABC} = \frac{1}{2} \times a \times b \times \sin C \]
Using the identity \( \sin C = 2 \sin\left(\frac{C}{2}\right) \cos\left(\frac{C}{2}\right) \):
\[ \text{Area}_{ABC} = \frac{1}{2} \times a \times b \times 2 \sin\left(\frac{C}{2}\right) \cos\left(\frac{C}{2}\right) = a \times b \times \sin\left(\frac{C}{2}\right) \cos\left(\frac{C}{2}\right) \]

7. **Equating the Two Expressions**: Set the two expressions for the area equal:
\[ 3(a + b) \sin\left(\frac{C}{2}\right) = a \times b \times \sin\left(\frac{C}{2}\right) \cos\left(\frac{C}{2}\right) \]

8. **Dividing by \( \sin\left(\frac{C}{2}\right) \)**: Assuming \( \sin\left(\frac{C}{2}\right) \neq 0 \):
\[ 3(a + b) = ab \cos\left(\frac{C}{2}\right) \]

9. **Substituting \( \cos\left(\frac{C}{2}\right) \)**: We know \( \cos\left(\frac{C}{2}\right) = \frac{1}{3} \):
\[ 3(a + b) = ab \cdot \frac{1}{3} \]
Multiplying through by 3:
\[ 9(a + b) = ab \]

10. **Rearranging**: Rearranging gives:
\[ ab - 9a - 9b = 0 \]

11. **Using the Relationship**: We can express this in terms of \( \frac{1}{a} + \frac{1}{b} \):
\[ \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \]
From \( ab = 9(a + b) \):
\[ \frac{1}{a} + \frac{1}{b} = \frac{a + b}{9(a + b)} = \frac{1}{9} \]

12. **Final Result**: Thus, the value of \( \frac{1}{a} + \frac{1}{b} \) is:
\[ \frac{1}{a} + \frac{1}{b} = 9 \]