Why is $ \arccos\left(\frac{1-x^2}{1+x^2}\right)= -2\arctan(x)$ true $\forall x\in (-\infty, 0]$? I'm trying to solve the problem that follows, and I'd appreciate any feedback on my solution in order to improve it. Thank you.
$$ \arccos\left(\frac{1-x^2}{1+x^2}\right) = -2\arctan(x) $$
So, I start off with analysing the domain. It's clear that arctan is defined for all values of x that are a member of the real numbers.
We can also see that the argument of arccos is less than 1 and greater than -1 for all x's. So that means, our equation is defined for all x's that belongs to the real numbers.
$$\Rightarrow \tan(\arccos(\frac{1-x^2}{1+x^2}) = \tan(-2\arctan(x))$$
Using the fact that $\tan(x) = \sqrt{1-\cos^2(x)}/\cos(x)$ as well as $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$. I land at this equation:
$$\frac{|2x|}{1-x^2} = \frac{-2x}{1-x^2}$$
Which is true for all negative x's, so basically $\forall x\in (-\infty, 0]$
I'm now asking you whether I can deduce this is true. Because, I actually applied tan to both sides, so does this imply that the solution set also holds for our original equation, and is there a way to deduce this, maybe using estimations?
Thank you.
 A: We have that the range for $\arctan x$ is $\left(-\frac \pi 2,\frac \pi 2\right)$ and therefore the range for $-2\arctan x$ is $\left(-\pi ,\pi \right)$ but the range for $\arccos\left(\frac{1-x^2}{1+x^2}\right)$ is $\left[0 ,\pi \right]$ therefore equality can possibly hold only for the range $\left[0,\pi \right)$ which implies $x\in(-\infty,0]$.
A: Starting from your equation $\def\acos{\operatorname{acos}}$
$\def\atan{\operatorname{atan}}$
$$\acos\frac{1-x^2}{1+x^2} = -2\atan x \tag{1}$$
And taking $\cos$ on both sides we get
$$(1)\ \implies\ \frac{1-x^2}{1+x^2} = \cos (2\atan x) \tag{2}$$
Notice that it's only an implication, and for what you are trying to show you'll need the other direction, so you must be carful and check whether it's an equivalence or not, or for which $x'$s it's an equivalence.
One common problem in such treatments can be that the trigonometric functions do not have inverses; they do only have inverse functions when restrict to, say, small enough intervals.  This is usually referred to as "taking a specific branch of a function".  For example, the usual choice for $\acos$ is a branch that satisfies $\acos x\in[0,2\pi]$, however it's totally legal, and depending on your problem might even be better, to have $\acos x\in[-\pi,\pi]$ for example.  Of cource you'll have to make clear in a paper or in a software which branch you are using when you are deviating from common notation.
To proceed with (2), observe that
$$\tan^2x
= \frac{\sin^2x}{\cos^2x}
= \frac{1-\cos^2x}{\cos^2x}
\qquad\text{thus}\qquad
\cos^2x = \frac1{1+\tan^2x}
\tag{3}$$
Substituting $x\mapsto \atan x$ turns (3) into
$$ \cos^2(\atan x) = \frac1{1+x^2} \tag{4}$$
Using the well known $\cos(2x) = \cos^2x - \sin^2x$, and using $\sin^2 + \cos^2=1$ to get a formula similar to (4) for $\sin^2$, we can finally write (2) as:
$$
\frac{1-x^2}{1+x^2}
= \underbrace{\cos^2(\atan x)}_{\tfrac1{1+x^2}} - \underbrace{\sin^2(\atan x)}_{\textstyle 1-\frac1{1+x^2}}
$$
and we are done.
