How to solve $2\tan^{-1}\sqrt{x-x^2}=\tan^{-1}x+\tan^{-1}(1-x)$ $$2\tan^{-1}\sqrt{x-x^2}=\tan^{-1}x+\tan^{-1}(1-x)$$
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
