Step 1: Understanding the Concept:
We are given that the sides $a, b, c$ of a triangle are in Arithmetic Progression (A.P.). This gives us a specific relation between them.
We need to evaluate a product of half-angle tangents. We should use the formulas expressing half-angle tangents in terms of the sides of the triangle.
Step 2: Key Formula or Approach:
If $a, b, c$ are in A.P., then $2b = a + c$.
Half-angle formulas for tangent:
$\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$
where $s$ is the semi-perimeter, $s = \frac{a+b+c}{2}$.
Step 3: Detailed Explanation:
Let's express the product $\tan \frac{A}{2} \cdot \tan \frac{C}{2}$ in terms of sides:
\[ \tan \frac{A}{2} \cdot \tan \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \]
Combine the square roots:
\[ = \sqrt{\frac{(s-b)(s-c)(s-a)(s-b)}{s(s-a)s(s-c)}} \]
Cancel common terms $(s-a)$ and $(s-c)$ from numerator and denominator:
\[ = \sqrt{\frac{(s-b)^2}{s^2}} \]
Since $s = \frac{a+b+c}{2}$, we have $s>b$, so $s-b>0$. Thus, the square root simplifies to:
\[ = \frac{s - b}{s} \]
Now, use the given condition that $a, b, c$ are in A.P.:
\[ 2b = a + c \]
We know the semi-perimeter is:
\[ s = \frac{a + b + c}{2} \]
Substitute $a + c = 2b$ into the semi-perimeter formula:
\[ s = \frac{(a + c) + b}{2} = \frac{2b + b}{2} = \frac{3b}{2} \]
Now substitute $s = \frac{3b}{2}$ into our simplified expression for the product:
\[ \frac{s - b}{s} = \frac{\frac{3b}{2} - b}{\frac{3b}{2}} \]
\[ = \frac{\frac{3b - 2b}{2}}{\frac{3b}{2}} \]
\[ = \frac{\frac{b}{2}}{\frac{3b}{2}} \]
The $b/2$ terms cancel out:
\[ = \frac{1}{3} \]
Step 4: Final Answer:
The value is $\frac{1}{3}$.