Expand the first term: $\frac{\tan A-\tan B}{\tan A(1+\tan A\tan B)}+\frac{{\sin }^{2}C}{{\sin }^{2}A}=1$.

Convert $\frac{{\sin }^{2}C}{{\sin }^{2}A}=\frac{{\tan }^{2}C}{{\tan }^{2}A}\cdot \frac{1+{\tan }^{2}A}{1+{\tan }^{2}C}$, equivalently $\frac{1+{\cot }^{2}A}{1+{\cot }^{2}C}$.

Let $\tan A=x$, $\tan B=y$, $\tan C=z$. The equation becomes

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

Clearing denominators: $x(x-y)(1+z^2)+(1+x^2)z^2(1+xy)=x^2(1+z^2)(1+xy)$.

Expanding and simplifying (collect terms in $z^2$ and constants) reduces to $z^2=xy$, i.e. ${\tan }^{2}C=\tan A\cdot \tan B$.

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

Option (1).