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

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

Put $\tan A = x,; \tan B = y,; \tan C = z$

$ \Rightarrow \frac{x - y}{(1 + xy)x} + \frac{(x^2 + 1)z^2}{x^2(z^2 + 1)} = 1 $

$ \Rightarrow x(x - y)(z^2 + 1) + z^2(1 + x^2)(1 + xy) = (1 + xy)x^2(1 + z^2) $

after solving we get

$ z^2 = xy \quad \therefore 1 + x^2 \ne 0 $

$ \therefore \tan^2 C = \tan A \cdot \tan B $

$ \Rightarrow \tan A,; \tan C,; \tan B$ are in G.P.