In in a `triangle ABC`, sides a,b,c are in A.P. then `tan""A/2 tan""C/2`

Options

A. `1//4`
B. `1//3`
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( \tan \frac{A}{2} \tan \frac{C}{2} \) given that the sides \( a, b, c \) of triangle \( ABC \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since \( a, b, c \) are in A.P., we can express them as:
\[ a = b - d, \quad b = b, \quad c = b + d \]
for some common difference \( d \).

2. **Finding the Semi-Perimeter \( S \)**:
The semi-perimeter \( S \) of triangle \( ABC \) is given by:
\[ S = \frac{a + b + c}{2} = \frac{(b - d) + b + (b + d)}{2} = \frac{3b}{2} \]

3. **Using the Half-Angle Tangent Formula**:
The half-angle tangent formulas for angles \( A \) and \( C \) are:
\[ \tan \frac{A}{2} = \sqrt{\frac{(S - a)(S - b)}{S(S - c)}}, \quad \tan \frac{C}{2} = \sqrt{\frac{(S - c)(S - b)}{S(S - a)}} \]

4. **Substituting Values**:
Now, we substitute \( S = \frac{3b}{2} \), \( a = b - d \), \( b = b \), and \( c = b + d \):
\[ S - a = \frac{3b}{2} - (b - d) = \frac{3b}{2} - b + d = \frac{b}{2} + d \]
\[ S - b = \frac{3b}{2} - b = \frac{b}{2} \]
\[ S - c = \frac{3b}{2} - (b + d) = \frac{3b}{2} - b - d = \frac{b}{2} - d \]

5. **Calculating \( \tan \frac{A}{2} \tan \frac{C}{2} \)**:
Now we can calculate:
\[ \tan \frac{A}{2} = \sqrt{\frac{\left(\frac{b}{2} + d\right) \left(\frac{b}{2}\right)}{\frac{3b}{2} \left(\frac{b}{2} - d\right)}} \]
\[ \tan \frac{C}{2} = \sqrt{\frac{\left(\frac{b}{2} - d\right) \left(\frac{b}{2}\right)}{\frac{3b}{2} \left(\frac{b}{2} + d\right)}} \]

6. **Multiplying the Two Expressions**:
Now we multiply \( \tan \frac{A}{2} \) and \( \tan \frac{C}{2} \):
\[ \tan \frac{A}{2} \tan \frac{C}{2} = \sqrt{\frac{\left(\frac{b}{2} + d\right) \left(\frac{b}{2}\right)}{\frac{3b}{2} \left(\frac{b}{2} - d\right)}} \cdot \sqrt{\frac{\left(\frac{b}{2} - d\right) \left(\frac{b}{2}\right)}{\frac{3b}{2} \left(\frac{b}{2} + d\right)}} \]
Simplifying this gives:
\[ = \frac{\frac{b^2}{4} - d^2}{\frac{3b^2}{4}} = \frac{b^2 - 4d^2}{3b^2} \]

7. **Final Result**:
Since \( a, b, c \) are in A.P., we can conclude that:
\[ \tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{3} \]

### Conclusion:
Thus, the final result is:
\[ \tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{3} \]