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Find possible value of $x$ such that

$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$

Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.

But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?

As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.

Átila Correia
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Orion_Pax
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3 Answers3

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I advise you to apply the identity

\begin{align*} \tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - \tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)} \end{align*}

so that you arrive at the following equation: \begin{align*} \frac{3x - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3 & \Longleftrightarrow \frac{x^{3} - 4x}{3x^{2} - 2} = 3\\\\ & \Longleftrightarrow x^{3} - 9x^{2} - 4x + 6 = 0\\\\ & \Longleftrightarrow (x^{3} + x^{2}) - 10(x^{2} + x) + 6(x + 1) = 0\\\\ & \Longleftrightarrow x^{2}(x^{2} + 1) - 10x(x + 1) + 6(x + 1) = 0\\\\ & \Longleftrightarrow (x^{2} - 10x + 6)(x + 1) = 0 \end{align*}

Hence the solution set is a subset of $S = \{5\pm\sqrt{19},-1\}$.

EDIT

It remains to determine if there are extraneous roots.

In order to do so, you can check out WolframAlpha (for example).

Átila Correia
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  • +1 That looks like a good answer, but that is barely a hint! – Theo Bendit Mar 22 '22 at 00:36
  • @TheoBendit thanks for the comment. Should I remove the hint title? – Átila Correia Mar 22 '22 at 00:37
  • Actually book method was like i did taking tan^-13 - tan^-1y and tan^-1(x-1) + tan^-1(x+1) apply formula both sides – Orion_Pax Mar 22 '22 at 00:38
  • @Orien_Pax (I guess) the method proposed in the book is based on the assumption the student does not know the $\tan$ identity of the sum of three arcs. If that is indeed the case, you can proceed as you suggested as well. – Átila Correia Mar 22 '22 at 00:40
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    @ÁtilaCorreia I'd just finish it off, personally. You're almost there, and the final steps are routine. If I were writing it for the first time, I'd probably end it at $$\frac{3x - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3,$$ or possibly just the $\tan$ angle sum identity, and call it a hint. But that's me, and I tend to have different standards to the community at large. :-) – Theo Bendit Mar 22 '22 at 00:41
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    @TheoBendit I will take your comment into consideration and finish the exercise. Thanks for the feedback by the way :) – Átila Correia Mar 22 '22 at 00:43
  • But my doubt is the cases of xy>1<-1 etc as the book consider only one of them – Orion_Pax Mar 22 '22 at 00:45
  • Shouldnt we need to check for all the possibilities in that tan^-1 x+- tan^-1 y formula? – Orion_Pax Mar 22 '22 at 00:47
  • I see okay so steps being arrived same in both is why you considered this okay fine – Orion_Pax Mar 22 '22 at 00:56
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    No, it's not equivalent when you take the tangent of both sides: it could introduce extraneous solutions where the difference between the two sides is a multiple of $\pi$. – Daniel Schepler Mar 22 '22 at 00:56
  • Just one more thing putting x= -1 and 5+√19 seems like not satisfying the equation can you check ? – Orion_Pax Mar 22 '22 at 00:57
  • Exactly @DanielSchepler i was thinking that only , it was just a coincidence in this problem otherwise not true , how will you solve if you go by my method , will you take all possible cases where the book dealt with just one case ? – Orion_Pax Mar 22 '22 at 00:58
  • Well, if you do have a solution, then taking tangents of both sides will still result in a correct equation, so $x\in { -1, 5 \pm \sqrt{19} }$. Then from there, you can plug in, or at least do a rough estimate of the arctangents which is accurate enough to determine whether the difference between the two sides in the original equation is 0, $\pi$, $-\pi$, $2\pi$, $-2\pi$, etc. – Daniel Schepler Mar 22 '22 at 01:02
  • @ÁtilaCorreia check my previous comment – Orion_Pax Mar 22 '22 at 01:11
  • @DanielSchepler is there any other method we can apply for this problem ? And my method of using tan^-1 (x+1) - tan^-1 (x-1) = tan^-3 - tan^-1 x and applying tan^-1x +-tan^-1(y) formula both sides is too lengthy because we need to take many cases like x^2-1><1 and 3x><-1 right? – Orion_Pax Mar 22 '22 at 01:14
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You do not need to consider cases.

Think first that you are looking for the zeros of function $$f(x)=\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) -k$$ the first derivative $$f'(x)=\frac{1}{1+(x+1)^2}+\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}=\frac {3 x^4+6 x^2+8 } {\left(x^2+1\right) \left(x^2-2 x+2\right) \left(x^2+2 x+2\right) }$$ is always positive; so, whatever is $k$, there is only one root.

$$k=\tan^{-1}(3)\implies f(0)=-\tan^{-1}(3)<0\quad \text{and}\quad f(1)=\frac{\pi }{4}-\cot ^{-1}(7)>0$$ So, the solution is $\in (0,1)$.

If you do not want to make anything complex, expand $f(x)$ as a series around $x=0$

$$f(x)=-\tan ^{-1}(3)+2 x-\frac{x^3}{6}+O\left(x^5\right)$$ Using the expansion to $O\left(x^3\right)$, the simplest estimate is $$x_0 \sim \frac 12 \tan ^{-1}(3) =0.624523$$ while the exact solution is $5-\sqrt{19}=0.641101$.

If you already now about series reversion, using the expansion to $O\left(x^5\right)$ gives $$x=\frac{1}{2} \left(f(x)+\tan ^{-1}(3)\right)+\frac{1}{96} \left(f(x)+\tan ^{-1}(3)\right)^3+O\left(\left(f(x)+\tan ^{-1}(3)\right)^5\right)$$ and since we want $f(x)=0$, then the estimate $$x=\frac{1}{96} \tan ^{-1}(3) \left(48+\tan ^{-1}(3)^2\right)=0.644821$$

Claude Leibovici
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Since $$\tan\big(\arctan x+\arctan y\big)\equiv\frac{x+y}{1-xy}$$ and $\arctan$ has principal range $\left(-\frac\pi2,\frac\pi2\right),$ thus, for each $(x,y)$ and some $k\in\{-1,0,1\},$ $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)+k\pi.$$

Therefore, \begin{align}&\arctan(x+1) + \arctan(x) + \arctan(x-1) = \arctan(3)\\ \implies&\arctan\left(\frac{x^3-4x}{3x^2-2}\right)=\arctan3+n\pi\quad\text{for some }n\in\{-2,-1,\ldots,2\}\\ \implies&\frac{x^3-4x}{3x^2-2}=3\\ \implies&\cdots\\ \implies&x=-1\:\:\text{or}\:\:5\pm\sqrt{19}\end{align}

$5-\sqrt{19}$ is the only non-extraneous solution. Desmos check.

ryang
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