Trigonometric systems of equations I seek to express $DE$ and $UI$ and $IE$ in terms of $\theta$:
$$\sec { \theta  } =DE+IE+UI \tag{1.1}\label{myeqone}$$
$$\frac { 1 }{ 4 } =UI(UI+IE) \tag{2.1}\label{myeqtwo}$$
$$(\frac { 1 }{ 2 } -\tan { \theta  } )^{ 2 }=DE(DE+IE) \tag{3.1}\label{myeqthree}$$
I rewrote $(2)$ and $(3)$ as
$$\frac { 1 }{ 4 } =UI(sec\theta -DE) \tag{2.2}\label{myeqtwo_one}$$
$$(\frac { 1 }{ 2 } -{ tan }^{ 2 }\theta )=DE(sec\theta -UI) \tag{3.2}\label{myeqthree_one}$$
This is where I'm stuck. I tried solving the system by replacing $DE$ in $(2.2)$ with $$\frac { (\frac { 1 }{ 2 } -tan^{ 2 }\theta ) }{ (sec\theta -UI) } $$but to no avail. I've also tried multiplying, adding, and subtracting the two systems in the hopes of something symmetric or meaningful, but once again to no avail. I even replaced $UI$, $DE$, and $IE$ with $x$,$y$, and $z$ in the hopes of simplifying some notation and hopefully solving the equation, but to no avail.
 A: Running this through Mathematica, I get ...
$$\begin{align}
IE &= \pm\sqrt{\sin 2\theta} \\[4pt]
UI &= \phantom{\sec\theta}-\frac12 \left(\pm\sqrt{\sin2\theta}-(\cos\theta+\sin\theta)\right)\\[4pt]
DE &= \sec\theta-\frac12\left(\pm \sqrt{\sin2\theta}+(\cos\theta+\sin\theta)\right) 
\end{align}$$
where the same "$\pm$" is used throughout.
The algebra for deriving the above looks a bit messy; if you really want to see it, I can take a shot at writing it down. Personally, I'd like to see how the equations arose; there may be a straightforward geometric argument that avoids most of the algebra.

Here's what I did in Mathematica ...
Mathematica doesn't always simplify Sec[] and Tan[] the way I like, so I tend to write equations in terms of cost and sint, and then do a separate simplification pass. In this case, the crux of the work is done via:
Solve[{
  1/cost == DE + IE + UI, 
  1/4 == UI (UI + IE), 
  (1/2 - sint/cost)^2 == DE (DE + IE)}, 
  {DE, IE, UI}]
Easy-peasy!
This gave two solutions (differing only in sign, hence the "$\pm$"), with this expression needing some attention:
Sqrt[cost^2 (-1 + sint^2) (-1 + cost^2 - 2 cost sint + sint^2)]
Without too much trouble, this simplifies to cost^2 Sqrt[2 sint cost]; that is, $\cos^2\theta \sqrt{\sin 2\theta}$. From there, a little clean-up gives the result as shown above.
A: For notational brevity, let $x=DE$, $y=IE$ and $z=UI$ and work with the set below,
$$x+y+z = \sec \theta \tag{1}$$
$$z(z+y) =\frac 14\tag{2}$$
$$x(x+y)=\left(\frac 12 -\tan\theta \right)^2\tag{3}$$
Take (3)-(2) and use (1),
$$x-z=\sin\theta(\tan\theta-1)\tag{4}$$
From (3)+(2),
$$x^2+z^2+y(x+z)=\left(\frac 12 -\tan\theta \right)^2+\frac14\tag{5}$$
Square (1) and rearrange,
$$y^2-(x-z)^2+2[x^2+z^2+y(x+z)]= \sec^2\theta\tag{6}$$
Substitute (4) and (5) into (6) to get the equation for $y$,
$$y^2-\sin^2\theta(\tan\theta-1)^2+2\left(\frac 12 -\tan\theta \right)^2+\frac12=\sec^2\theta$$
Then, simplify and obtain the solution for $y$,
$$y^2=\sin 2\theta,\>\>\>\>\>y=\pm\sqrt{\sin 2\theta}$$
Then, the solutions for $x$ and $z$ follow,
$$\begin{align}
x&= -\frac12\left(y+\cos\theta+\sin\theta\right) +\sec\theta\\ 
z&= -\frac12 \left(y-\cos\theta-\sin\theta\right)
\end{align}$$
