Solving $\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$ If we have to find the solutions of equation

$$\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$$

Using a triangle I rewrite it as
$$2 \arctan \left(\frac{\sqrt{1-x^2}}{x}\right)= 0$$
So this equation is satisfied when $x=\pm 1$ 
But I saw that $x=-1/2$ is also satisfying , then where I have missed the case . 
I am totally stuck , how to find it .
 A: For $x<0$ set $x=\sin z$. Then you have
$$\arcsin(\sqrt{1-x^2})=\arcsin(\cos z)= \frac{1}{2}\pi - \arccos(\cos z)$$
$$\arccos(\sin z)= \frac{1}{2}\pi - \arccos(\cos z)$$
$$\arcsin x= \arcsin (\sin z)= z$$
$$\mathrm{arccot}(\frac{\sqrt{1-x^2}}{x})=\mathrm{arccot}(\frac{\cos z }{\sin z})
=\mathrm{arccot}(\cot z) = \pi + z$$
(Note: The last expression would be $z$ for $z>0.$) Now compute the sums
$$\arcsin(\sqrt{1-x^2}) + \arccos x = \pi$$
$$\mathrm{arccot}(\frac{\sqrt{1-x^2}}{x})- \arcsin x = \pi$$
So both sides equal $\pi$ for $x<0$
A: Straightaway the problem reduces to  $$\text{arccot}\dfrac{\sqrt{1-x^2}}x=\dfrac\pi2+\arcsin\sqrt{1-x^2}$$ 
As $\sqrt{1-x^2}\ge0,$ using  the definition of Principal Values
 $0\le\arcsin\sqrt{1-x^2}\le\dfrac\pi2$
and consequently, $\dfrac\pi2\le\text{arccot}\dfrac{\sqrt{1-x^2}}x\le\pi$
$\implies x\not>0$ but $x\ne0,$
let $-x=y>0$
$$\implies\text{arccot}\dfrac{\sqrt{1-y^2}}{-y}=\dfrac\pi2+\arcsin\sqrt{1-y^2}$$
$$\iff\dfrac\pi2-\arctan\dfrac{\sqrt{1-y^2}}{-y}=\dfrac\pi2+\arccos y$$
As $\arctan(-u)=-\arctan u,$
$$\arctan\dfrac{\sqrt{1-y^2}}y=\arccos y$$
Now as $y>0$ and let $\arccos y=v\implies\cos v=y$ and $0\le v<\dfrac\pi2$
and $\dfrac{\sqrt{1-y^2}}y=\tan v\implies\arctan\dfrac{\sqrt{1-y^2}}y=v=\arccos y$ as $0\le v<\dfrac\pi2$ 
So, we need $y>0\iff x<0$
