If $$\frac{\tan (A-B)}{\tan A}+\frac{\sin^{2}C}{\sin^{2}A}=1,A,B,C \in \left(0,\frac{\pi}{2}\right)$$, Then

Given: $$\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1$$

1. Rearrange: $$\frac{\sin^2 C}{\sin^2 A} = 1 - \frac{\tan(A-B)}{\tan A} = 1 - \frac{\sin(A-B)\cos A}{\cos(A-B)\sin A}$$
2. Simplify RHS: $$\frac{\sin A \cos(A-B) - \cos A \sin(A-B)}{\sin A \cos(A-B)} = \frac{\sin(A - (A-B))}{\sin A \cos(A-B)} = \frac{\sin B}{\sin A \cos(A-B)}$$
3. Equate: $$\frac{\sin^2 C}{\sin^2 A} = \frac{\sin B}{\sin A \cos(A-B)} \implies \sin^2 C = \frac{\sin A \sin B}{\cos A \cos B + \sin A \sin B}$$
4. Divide numerator and denominator by $$\cos A \cos B$$:
5. Since $$\sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C}$$, by comparison: $$\tan^2 C = \tan A \tan B$$

$$\sin^2 C = \frac{\tan A \tan B}{1 + \tan A \tan B}$$

Conclusion: $$\tan A, \tan C, \tan B$$ are in G.P. (Option D)