Regarding an inverse trigonometric equation. I tried to find the solutions of this equation
$$
\arctan\left(\frac{2x}{1-x^2}\right)+\text{arccot}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi}{3}
$$
I got solutions $\frac{1}{\sqrt{3}}$ and $-\sqrt3$. by reciprocating the arc cot term into arc tan term, adding both and solving the equation. But in solutions, in addition to above answers, $\sqrt3+2$ and $\sqrt3-2$ has also been given as answers, which I cannot figure out how they came?
 A: Please note that:$\arctan\left(\frac{2x}{1-x^2}\right)+\text{arccot}\left(\frac{1-x^2}{2x}\right)$=\begin{cases}
2\arctan\left(\frac{2x}{1-x^2}\right),  & \text{if $\left(\frac{2x}{1-x^2}\right)$$\gt$$0$} \\
2\arctan\left(\frac{2x}{1-x^2}\right)+\pi, & \text{if $\left(\frac{2x}{1-x^2}\right)$ $\lt$$0$}
\end{cases}. 
In the first case, you get x=1/$\sqrt3$,x=-$\sqrt3$.
In the second case, you get x=$\sqrt3$+2,$\sqrt3$-2.
P.S.  I am using principal values for all inverse trigonometric ratios involved in here.
A: Use:
Let $\cot^{-1}x=\theta$ 
$x=\cot\theta$ 
$$\boxed{\tan^{-1}\frac1x=\theta}$$ 
$$\boxed{\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}} \ \ \text{(Not necessarily used)}$$
So $$\cot^{-1}\frac{1-x^2}{2x}=\tan^{-1}\frac{2x}{1-x^2}$$
$$\tan^{-1}\frac{2x}{1-x^2}=\frac{\pi}3$$
$$\frac{2x}{1-x^2}=\sqrt3$$
$$2x=\sqrt3-\sqrt3x^2$$
A: EDIT: Answer revised (explaining the difference in results) in light of comments. (And I really can't take credit for this, as it was other people who pointed out the variation in the definition of the inverse trigonometric functions.)
If you try this on Wolfram Alpha, for example, it will plot the arc cotangent function like this:

Therefore if you try substituting either the "solution" $\sqrt 3 + 2$ for $x$ in $\text{arccot}\left(\frac{1-x^2}{2x}\right)$, Wolfram Alpha evaluates it as $-\frac\pi6$.
You will get the same result if you substitute $\sqrt 3 - 2$
for $x$.
This explains why one might reasonably not see $x = \sqrt3 \pm 2$
as a solution of the equation.
But evidently the person writing the answer key for this problem takes the definition of the arc cotangent function that has $\frac\pi2<\cot^{-1}(x)<\pi$
for all $x < 0$, as do many other people (a definition that is reflected in various mathematical software tools).
In that case the arc cotangent term in the question evaluates to $\frac56\pi$
rather than $-\frac\pi6$
when $x = \sqrt3 \pm 2$, thereby solving the equation.
