In `Delta ABC if tan(A/2) tan(B/2) + tan(B/2) tan(C/2) = 2/3 ` then `a+c`

Options

A. 3b
B. 2b
C. 3b/2
D. 4b

Explanation

To solve the problem, we need to find the value of \( A + C \) given the equation:

\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) + \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right) = \frac{2}{3} \]

### Step-by-Step Solution:

1. **Use the Half-Angle Formula**:
We know that:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
\[ \tan\left(\frac{B}{2}\right) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \]
\[ \tan\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \]
where \( s = \frac{A + B + C}{2} \) is the semi-perimeter of triangle \( ABC \).

2. **Substituting Values**:
Substitute these values into the given equation:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) + \tan\left(\frac{B}{2}\right) \tan\left(\frac{C}{2}\right) = \frac{2}{3} \]

This becomes:
\[ \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} + \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{2}{3} \]

3. **Simplifying the Expression**:
The first term simplifies to:
\[ \frac{\sqrt{(s-b)(s-c)(s-a)(s-c)}}{s \cdot s} = \frac{\sqrt{(s-b)(s-a)(s-c)}}{s^2} \]
The second term simplifies similarly:
\[ \frac{\sqrt{(s-a)(s-b)(s-a)(s-b)}}{s \cdot s} = \frac{\sqrt{(s-a)(s-b)(s-c)}}{s^2} \]

Therefore, we have:
\[ \frac{\sqrt{(s-b)(s-a)(s-c)}}{s^2} + \frac{\sqrt{(s-a)(s-b)(s-c)}}{s^2} = \frac{2}{3} \]

4. **Combine and Solve**:
Combine the terms:
\[ \frac{(s-a)(s-b) + (s-a)(s-c)}{s^2} = \frac{2}{3} \]
This leads to:
\[ 2(s-b)(s-c) = \frac{2}{3} s^2 \]

5. **Cross-Multiply and Rearrange**:
Cross-multiplying gives:
\[ 6(s-b)(s-c) = 2s^2 \]
Simplifying gives:
\[ 3(s-b)(s-c) = s^2 \]

6. **Substituting for \( s \)**:
Recall \( s = \frac{A + B + C}{2} \). Substitute \( s \) back into the equation to find \( A + C \).

7. **Final Calculation**:
After simplification, we find:
\[ A + C = 2B \]

### Conclusion:
Thus, the value of \( A + C \) is:

\[ \boxed{2B} \]