find the tanx

Let $CF=DE =CE = 2$ and $GH$ be perpendicular to $CD$ . Then $\triangle CEF$ and $\triangle EGH$ are similar. Therefore $\dfrac {GH}{CF} = \dfrac {EG}{GH} = \dfrac 12$ , $\implies GH = \dfrac 12 \times CF = 1 = CE$ . Then we have:

$\begin{aligned} \angle DGE & = \angle DGH - \angle EGH \\ x & = \tan^{-1} \frac {DH}{GH} - \tan^{-1} \dfrac {EH}{GH} \\ & = \tan^{-1} 3 - \tan^{-1} 1 \\ \implies \tan x & = \frac {3-1}{1+3} = \frac 12 = \boxed {0.5} \end{aligned}$

Without loss of generality, let the length of the side of the square be 2 units. Then, EC = FC = DE = 1. By the Pythagorean Theorem applied to triangle ACF, EF = sqrt(2) so that GF = EG = 1/sqrt(2). Also, angle FEC = 45 degrees, so that angle DEG = 180 - 45 = 135 degrees. Applying the law of cosines to triangle DGE yields:

DG^2 = 1 + 1/2 - 2 (1) (1/sqrt(2))*cos(135), which means:

DG = sqrt(5/2).

Now, applying the law of sines to DEG yields:

sin(x) / 1 = sin(135) / DG, or

sin(x) = 1/sqrt(5).

Therefore, by right triangle trigonometry (or by trig identities), tan(x) = 1/2