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.)

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.
