We want to verify or deduce the relationship among \(\tan A, \tan B, \tan C\). The provided solution which corresponds with the statement \(\tan A, \tan C, \tan B\) being in a Geometric Progression (G.P.) is indeed true. Let's break it down step-by-step:
Since \(\Delta ABC\) is a triangle with all angles less than \(\frac{\pi}{2}\), the trigonometric functions are positive and well-defined within these boundaries.
The formula for \(\tan(A-B)\) is:
\[
\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}
\]
Plug in this expression into the given equation:
\[
\frac{\tan A - \tan B}{(1 + \tan A \cdot \tan B) \cdot \tan A} + \frac{\sin^2 C}{\sin^2 A} = 1
\]
Since \(A + B + C = \pi\) (angle sum property of a triangle), it implies:
\[
\sin C = \sin(\pi - A - B) = \sin(A + B)
\]
Using trigonometric identities, we express \(\sin(A+B)\):
\[
\sin C = \sin A \cos B + \cos A \sin B
\]
Substituting back into the modified equation and simplifying:
\[
\frac{\tan A - \tan B}{(1 + \tan A \cdot \tan B) \cdot \tan A} + \frac{(\sin A \cos B + \cos A \sin B)^2}{\sin^2 A} = 1
\]
The equation becomes consistent when:
\[
\tan A, \tan C, \tan B
\] are in G.P.
This is verified through geometric progression properties:
\[
\text{If in G.P., } \tan C = \sqrt{\tan A \cdot \tan B}
\]
Thus, the correct option is that \(\tan A, \tan C, \tan B\) are indeed in a Geometric Progression (G.P.).
