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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?

SHREE6174
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  • By $\tan^{-1}(x)$, do you mean $\arctan(x)$ or $\frac{1}{\tan(x)}$? – 5xum Nov 05 '14 at 14:19
  • It's $\arctan$. Inverse trigonometric – SHREE6174 Nov 05 '14 at 14:21
  • see http://math.stackexchange.com/questions/443655/prove-that-arctan-left-frac2x1-x2-right-2-arctanx-for-all-x1-d, there is a really nice argument from Andre Nicolas – Mr. Barrrington Nov 05 '14 at 14:53
  • @SHREE6174, See http://math.stackexchange.com/questions/138310/show-that-2-tan-12-pi-cos-1-frac35/583359#583359 – lab bhattacharjee Nov 05 '14 at 15:49
  • @labbhattacharjee For $tan^{-1}x+tan^{-1}y=tan^{-1}((x+y)/(1-xy))$ Why $xy<1$? How can we prove it? – SHREE6174 Nov 05 '14 at 16:35
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    @SHREE6174, Have you noticed the two links in my answer, there? – lab bhattacharjee Nov 05 '14 at 16:41
  • Check my answer, I've updated it and provided you with the detailled solution. I hope I've been clear, it's always tricky when manipulating the inverse of tan because of the restriction you have to put on the argument of tan ( only things between $-\pi$/2 and $\pi$/2). – mvggz Nov 10 '14 at 14:24

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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.

mvggz
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  • I am not getting. I have plotted separate graphs for this problem. I noticed that $2tan^{-1}(x)$ is continuous curve and range is $(-\pi,\pi)$, but Right hand side is not, it is discontinuous at -1 and 1 with range $(-\pi/2,\pi/2)$. Both curves are same at $-1<x<1$.How can we conclude without graph? – SHREE6174 Nov 05 '14 at 16:29
  • Use what I told you to prove both expression are equal on ]-1;1[. You evaluate the tan of 2arctan(x) with the formula giving tan(2a) and you will find what you want. As for the rest, since both expression are odd funtions you can only study the case x>1. Here try using : $arctan(x) + arctan(\frac{1}{x}) = \frac{\pi}{2}$ to bring the expression back into values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ – mvggz Nov 05 '14 at 21:52