Question

$\frac{tanA - tanB}{(1 + tanAtanB)tanA} + \frac{1 + \cot^{2}\text{ }A}{1 + \cot^{2}C} = 1$

Put $tanA = x,tanB = y,tanC = z$

$${\therefore\frac{x - y}{(1 + xy)x} + \frac{\left( x^{2} + 1 \right)z^{2}}{x^{2}\left( z^{2} + 1 \right)} = 1 }{\therefore x(x - y)\left( z^{2} + 1 \right) + z^{2}\left( 1 + x^{2} \right)(1 + xy) }{= (1 + xy)x^{2}\left( 1 + z^{2} \right) }$$after solving we get

$${z^{2} = xy }{\therefore\tan^{2}C = tanA \cdot tanB }$$$tanA,tanC,tanB$ are in G.P.