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$\begingroup$

$$ \sqrt{\arctan(x)} = \dfrac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right)$$

I have been trying to solve this problem for the past hour, but I'm not able to solve it as I have just started solving difficult trigonometric problems. I'm not able to get any logic to solve this problem. I'm not able to put any trigonometric formula to solve this please help.

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    This looks like a math problem, nothing to do with Mathematica. Migrate?2012-08-27
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    This statement is not true: `Sqrt[ArcTan[x]] === 1/2 ArcCos[(1 - x)/(1 + x)]` --> `False`. Can anybody check if my TeX-ification is correct?2012-08-27
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    @stevenvh the original `1-x/1+x` could be interpreted variously as `(1 - x)/(1 + x)`, `1 - x/(1 + x)`, or `1`. However, none of these make the statement true (except for $x=0$). So I see no problem with your interpretation but still don't really know how to proceed.2012-08-27
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    this is a site for questions on [Mathematica](http://www.wolfram.com/mathematica/), not mathematics (also, the two sides of your equation aren't actually equal).2012-08-27
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    This user already has several questions like this on Math.SE, so it's a bit strange to post this here now, having found the right place in the past. Even though it was purely accidental, hopefully they'll be glad that we saved them from a low-quality question...2012-08-27
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    It's not true. Try typing Plot[{Sqrt[ArcTan[x]], (1/2) ArcCos[(1-x)/(1+x)]}, {x,0, 10}] into Wolfram Alpha. Conceptually, you shouldn't expect this to be true: $\tan^{-1}(x)$ is an angle, and taking the square root of an angle has no physical meaning.2012-08-27
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    Taking limits, one can observe that $\lim_{x \to \infty}(\textrm {LHS}) = \sqrt{\pi / 2}$, while $\lim_{x \to \infty}(\textrm {RHS}) = \pi / 2$. This not only shows that there is an error, but also suggests where to look in order to fix it: the $\sqrt{}$ is misplaced (as David Speyer also noted).2012-08-27
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    I think you mean $\arctan \sqrt{x}$, not $\sqrt{\arctan x}$.2012-08-27
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    Do not modify surreptitiously your question after you received several answers. This makes said answers look off-topic and is contrary to the policy of the site. I reverted your question to its previous version.2012-08-29
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    Satisfied with an answer below?2012-09-19

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