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

Question

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 .

Answer 1

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 $y>0\iff x<0$

Answer 2

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 $x<0$