$$2\tan^{-1}\sqrt{x-x^2}=\tan^{-1}x+\tan^{-1}(1-x)$$ I wasn't sure on how to go from here because I tried to draw triangles for each tangent function but that didnt work and I know that I can't distribute via doing tangent on both sides.
(though I do enjoy complex methods in solving this, I do appreciate a high school level process of doing this problem.)
I assume you want to prove equality between both sides. I show that your equality is almost correct but off by a constant of $\pi/4$
This is easy enough with calculus. If we differentiate either side we get
$$\frac{2 - 4 x}{x^4 - 2 x^3 + x^2 + 1}$$
and thus our functions differ by at most a constant. Setting $x=0$ we find
$$2\arctan\sqrt{x-x^2}\bigg|_{x=0}=2\arctan(0)=0\\ [\arctan(x)+\arctan(1-x)]\bigg|_{x=0}=\arctan(0)+\arctan(1)=\frac{\pi}{4}$$
We thus find that we must add $\pi/4$ to the LHS of your equality to have both sides be equal
Letting $u = 1-x,v=x,$ you need to solve $2\arctan \sqrt{uv} = \arctan u + \arctan v.$
Taking tangents on both sides and applying tangent addition and double angle formulae gives $\frac{2\sqrt{uv}}{1-uv}=\frac{u+v}{1-uv}\implies 2\sqrt{uv}=u+v\implies 2\sqrt{x-x^2}=1\implies 4(x-x^2)=1$ which is a simple quadratic to solve.
HINT:
We need $x-x^2\ge0\iff 0\le x\le1$
Again, $x-x^2=\dfrac{1-(2x-1)^2}4\le\dfrac14$
$\implies0\le x-x^2\le\dfrac14$
Now use my answer here to find $x=\dfrac12$
