Inverse Trigonometry System of Equations

Question

$$2\tan^{-1}\left(\sqrt{x-x^2}\right) = \tan^{-1}\left(x\right)\: +\, \tan^{-1}\left(1-x\right)$$

I have a feeling solution includes drawing triangles but cannot make the leap to get the solution

Answer 1

Take tangent of both sides (apply double angle and sum of angle formula and remember that $\tan(\tan^{-1}y) = y$):

$$\frac{2\sqrt{x-x^2}}{1 - (x-x^2)} = \frac{x+1-x}{1-x(1-x)}$$

and solve that equation.

Answer 2

We need $x(1-x)\ge0\iff x(x-1)\le0\iff0\le x\le1$

$\implies\dfrac{x+1-x}2\ge\sqrt{x(1-x)}$

Now use $$\arctan x+\arctan y= \arctan\frac{x+y}{1-xy}$$ if $xy<1$

See Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

OR showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$