Tangents?!

Since $A + B + C = 180, A + C = 180 - B,$ and $\tan(A+C) = \tan(180 - B) = -\tan(B).$ Expanding,

$\frac{\tan A + \tan C}{1 - \tan A \tan C}$ $= -\tan B.$ The tangents are in arithmetic progression, so

$\tan C - \tan B = \tan B - \tan A;$ so $\tan B =$ $\frac{\tan A + \tan C}{2}$ . Since $\tan A + \tan C$ is nonzero,

$-\frac{\tan A + \tan C}{1 - \tan A \tan C} = \frac{\tan A + \tan C}{2}$ . Simplifying, $\tan A \tan C$ = 3.

Since $\tan(A), \tan(B)$ and $\tan(C)$ from an arithmetic progression, let their values be $b-d, b$ and $b+d$ respectively for some reals, $b$ and $d$ and $b > 0$ . Using the property that, for a triangle with angles $A ,B$ and $C$ , $\tan(A) + \tan(B) + \tan(C) = \tan(A)\tan(B)\tan(C)$ , we can write,
$\displaystyle b - d + b +b + d = (b-d)b(b+d) \Rightarrow 3b = b (b^2 - d^2) \Rightarrow b^2 - d^2 = 3$ .

But, $\tan(A) \times \tan(C) = (b-d)(b+d) = b^2 - d^2$ . Thus the required answer is $\displaystyle \boxed{3}$ .