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Can someone help me simplify the argument of $\arctan$ in this problem ? $$\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)$$

C. Dubussy
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Hani
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4 Answers4

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It might be helpful to take the inverse of this function by solving for $x$.

We have the following: $$y=\arctan\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ Take the $\tan$ of both sides: $$\tan y=\frac{\sqrt{1+x^2}-1}{x}$$ Muultiply both sides by $x$ and add $1$: $$x\tan y+1=\sqrt{1+x^2}$$ Square both sides: $$x^2\tan^2 y+2x\tan y+1=1+x^2$$ Subtract both sides by $1+x^2$: $$x^2(\tan^2 y-1)+2x\tan y=0$$ Now, clearly, the original function is undefined for $x=0$, so we can conclude that $x \neq 0$. Therefore, divide by $x$: $$x(\tan^2 y-1)+2\tan y=0$$ Subtract both sides by $2\tan y$ and divide by $\tan^2 y-1$: $$x=-\frac{2\tan y}{\tan^2 y-1}$$ Now, this looks very similar to the $\tan(2y)$ identity. However, the denominator is flipped and there is a $-$ sign out front. Therefore, distribute the negative across the denominator: $$x=\frac{2\tan y}{1-\tan^2 y}=\tan 2y$$ Now, we can solve this equation for $y$ to get a much simpler form of our original equation. Take the $\arctan$ of both sides: $$\arctan x=2y$$ Divide both sides by $2$ and switch the sides of the equation: $$y=\frac{\arctan x}{2}$$

Noble Mushtak
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  • Thanks.... I got to know a new method! – Hani Jun 13 '16 at 18:41
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    Glad I could help! I'm not very good with trig identities and there's no way I would notice something like @Qwerty did in his answer, so taking the inverse of functions is what I usually do to solve these kinds of problems. – Noble Mushtak Jun 13 '16 at 18:42
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Take $x=\tan \theta$ So it becomes ${\sec\theta -1\over tan \theta}$ multiply up and down by $\cos\theta$. Becomes ${1-cos\theta\over sin \theta}={2\sin^2(\theta/2)\over 2\sin(\theta/2)\cos (\theta/2)}=\tan(\theta/2)$

So the answer is $\theta/2 = (tan^{-1}x)/2$

Qwerty
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Differentiate with respect to $x$: $$ \frac{\partial}{\partial x}\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)=\frac{\frac{1}{x^2+\sqrt{x^2+1}+1}}{1+\left(\arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)\right)^2}=\frac{1}{2(1+x^2)} $$ Thus: $$ \arctan\left(\frac{\sqrt{1 + x^2} -1}{x}\right)=\frac{1}{2}\arctan(x)+C $$ Since both sides are $0$ at $x=0$, then $C=0$.

asomog
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    "Since both sides are $0$ at $x=0$" $\rightarrow$ How is the left-side defined at $x=0$? There is an $x$ in the denominator. I think checking $x=\sqrt 3$ makes more sense. – Noble Mushtak Jun 13 '16 at 18:43
  • How both sides zero? – Hani Jun 13 '16 at 18:45
  • @NobleMushtak Except for $0$, quite trivially: $$ \frac{\sqrt{1 + x^2} -1}{x}=\frac{x}{(\sqrt{x^2+1}+1)} $$ Then take $x\to 0$. – asomog Jun 13 '16 at 18:52
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    @ÁkosSomogyi Oh, OK. That works, but I just think it's easier to explain with $x=\sqrt 3$ since you only need to plug the value in instead of manipulating the expression and taking the limit. – Noble Mushtak Jun 13 '16 at 18:55
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There's a geometric reason for why you get $\frac12 \arctan x$. Write your expression as $\arctan\frac{x}{\sqrt{1+x^2}+1}$ and look at the pictures in this answer (about taking square roots of complex numbers) for the number $z=1+ix$, so that $\phi=\arctan x$ and $r = \sqrt{1+x^2}$, and note that $w=(x+\sqrt{1+x^2})+ix$ so that your expression is the argument of $w$.

Community
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Hans Lundmark
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