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How to prove the following identity:

$$(\forall x\in\mathbb{R}): \quad \arctan(\sqrt{x^2+1}-x)=\dfrac{\pi}{4}-\dfrac{1}{2}\arctan(x)$$

The knowing method is to prove that both sides are in $]-\frac{\pi}{2}, \frac{\pi}{2}[$ and they have the same value tooked by the function $\tan$; but this looked to be out of reach, because there is no general identity giving $\tan(\frac{\alpha}{2})$ in words of $\tan(\alpha)$

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    $\begingroup$ "there is no general identity giving $\tan(\frac\alpha2)$ in words of $\tan(\alpha)$" ummmmm.... the half angle identities? $\endgroup$ – YiFan Oct 8 '19 at 23:40
  • $\begingroup$ @YiFan, I mean an identity giving $\tan(\frac{\alpha}{2})$ in terms of $ \tan(\alpha)$, this the only way that helps proving the equality. $\endgroup$ – hachemy Oct 8 '19 at 23:49
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    $\begingroup$ Have a look at these demonstrations, solely based on geometrical definition of tangent dfnu.xyz/en/exercises-and-dialogues/losing-it-on-a-tangent $\endgroup$ – dfnu Oct 9 '19 at 12:36
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Let $\theta=\arctan(\sqrt{x^2+1}-x),$ then $\tan\theta=\sqrt{x^2+1}-x$ and we can easily solve this for $x,$ as $$x=\dfrac{1-\tan^2\theta}{2\tan\theta}=\cot 2\theta=\tan\left(\dfrac{\pi}2-2\theta\right)$$ which implies $$\theta=\dfrac{\pi}4-\dfrac12\arctan(x).$$

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    $\begingroup$ Nice solution! Thank you @Bumblebee's $\endgroup$ – hachemy Oct 9 '19 at 0:01
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Hint

Let arccot$(x)=2y,x=\cot2y,0<2y<\pi$

$\sqrt{1+x^2}-x=+\csc2y-\cot2y=\tan y$

$\arctan(x)=\dfrac\pi2-$arccot$(x)=\dfrac\pi2-2y$

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