What are the possible values of $x$?

Question

For what values of $x$ does this equation holds? $$2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right)$$ The answer is $-1<x<1$ Why? How can we say this?

Answer 1

Hint :

$-1 < x < 1 => 2*arctan(x) \in]-\frac{\pi}{2};\frac{\pi}{2}[$ ; $\tan(2a) = \tan(a+a) = \frac{\tan(a)+\tan(a)}{1-\tan(a)*\tan(a)} = \frac{2\tan(a)}{1-\tan^2(a)}$

tan is bijective on $]-\frac{\pi}{2};\frac{\pi}{2}[$, calculate : tan(2*arctan(x)), see what you get.

Edit: I'll go a bit deeper into the solution.

First the two expression being odd functions, you can limit the study to x$\in [0;+\infty[$.

You note that if the two quantity are equals, there has to be a singularity at x=1 since the expression on the right $\rightarrow \frac{\pi}{2}$ when : $x \rightarrow 1 $ with $x<1$, and $\rightarrow -\frac{\pi}{2}$ when $x \rightarrow 1 $ with $x>1$. So it is not continuous at x=1 while the other expression is. We'll start by treating both cases seperately.

Case x >1:

g: $ x \rightarrow \frac{2x}{1-x^2} $ has negative values only, since $1-x^2 <0$

In fact : $ g(]1;+\infty[[) = ]-\infty;0[$ This means that: $\arctan(g(x)) < 0$, for $ x>1$ since arctan is of the same sign as its argument. But $2*\arctan(x) >0$ when x>1 (still of the same sign as its argument) so the two expressions can never be equal when x>1.

Case 0 < x < 1 :

Here I'm going to use the fact that $:u \rightarrow \tan(u)$ is bijective when $u\in]-\frac{\pi}{2};+\frac{\pi}{2}[$.

This means that if $u\in]-\frac{\pi}{2};+\frac{\pi}{2}[$, then $\arctan(\tan(u)) = u$

Let's evaluate $ \tan(2*\arctan(x))$, using the formula I gave you:

I'll let : $u= \arctan(x)$

$ \tan(2*\arctan(x)) = \tan[ u + u ] = \frac{2\tan(u)}{1-\tan^2(u)} $

Here $u=\arctan(x)$ , so : $\tan(u) = x$ and we end up with:

$ \tan(2*\arctan(x)) = \frac{2x}{1-x^2} $

0 < x < 1: $ a=2*\arctan(x) \in ]0;+\frac{\pi}{2}[ $ : As I said above we can write:

$ \arctan(\tan(a)) = a $ => $ 2*\arctan(x) = \arctan(\frac{2x}{1-x^2}) $

So you have proved that this formula is true for : 0 < x < 1, and therefore for: -1< x < 1 because of imparity. It is however false when x>1 or x<-1.