Find all $x$ that satisfy the equation $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$$


Taking tangent on both sides, and using the identities $\displaystyle \tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ and $\displaystyle \tan\left(2\alpha\right)=\frac{2\tan\alpha}{1-\tan^{2}\alpha}$,

We obtain $$\frac{\frac{\left(2x+1\right)}{x+1}+\frac{\left(2x-1\right)}{x-1}}{1-\frac{\left(4x^{2}-1\right)}{x^{2}-1}}=\frac{2\left(x+1\right)}{1-\left(x+1\right)^{2}}$$ and after removing the root $x=0$, we are left with a simple equation $(x^2-2)(2x+1)=0$. So, the possible roots are $x=0,\pm \sqrt{2}, -\dfrac{1}{2}$ , now my question is, how do I verify which roots satisfy the original equation, because one cannot see directly after plugging, whether it's true or not because it involves arctangents.

To avoid this situation, I re-wrote the original equation as $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)-\tan^{-1}\left(1+x\right)=\tan^{-1}\left(1+x\right)-\tan^{-1}\left(\frac{2x-1}{x-1}\right)$$ and because while using the identity $\tan^{-1}\left(\alpha\right)-\tan^{-1}\left(\beta\right)=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$, there are no restrictions on $\alpha,\beta$ I went ahead with this, and got the equation $$\frac{\frac{\left(2x+1\right)}{x+1}-\left(x+1\right)}{2\left(x+1\right)}=\frac{\left(x+1\right)-\frac{\left(2x-1\right)}{x-1}}{2x}$$ which simplified to $\left(x+1\right)^{2}\left(x-2\right)+x^{2}\left(x-1\right)=0$ and surprisingly from here, I am not even getting those roots!

I used this method to avoid the extraneous roots, what is going wrong here?

Edit: The immediately above issue has been resolved, however, the first issue remains unresolved.

Note that all calculations have to be performed by hand, since these types of questions are asked in exams in which the allotted time per question averages out to 2-3 minutes.

  • 1
    $\begingroup$ The $2x$ in your last denominator appears to be faulty. It looks like you cancelled $1+x$ with $x-1$. $\endgroup$ Apr 3, 2021 at 10:00
  • $\begingroup$ @GerryMyerson Oh yes! Thanks for pointing out. In the meantime, I also discovered that with the second approach, one needs to use $\alpha \beta >-1$ which I didn't consider. However, how do I verify the solutions with the first approach? Or How do I get the exact solutions directly from some method... $\endgroup$
    – V.G
    Apr 3, 2021 at 10:03
  • $\begingroup$ On verifying which roots are correct solutions: tan is a bijection on its domain onto $\mathbb R$, so the only thing you need to check is whether any of said solutions falls outside the domain of the various quantities involved, and then you are good to go. $\endgroup$
    – user145413
    Apr 3, 2021 at 10:03
  • $\begingroup$ Two preliminary observations: 1. " So, the possible roots are $x=0,\pm \sqrt{2}, -\dfrac{1}{2}$ , now my question is, how do I verify which roots satisfy the original equation (because one cannot see directly after plugging, whether it's true or not because it involves arctangents)? " A straightforward plug-and-check reveals that $x=0,\sqrt{2},-\dfrac{1}{2}$ satisfy the given equation, whereas $x=-\sqrt{2}$ doesn't. Why was the arctangent function an obstruction? $\endgroup$
    – ryang
    Apr 3, 2021 at 10:09
  • $\begingroup$ 2. " while using the identity $\tan^{-1}\left(\alpha\right)-\tan^{-1}\left(\beta\right)=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$, there are no restrictions on $\alpha,\beta$ " This identity may not be true when $\alpha\beta\nless 1.$ $\endgroup$
    – ryang
    Apr 3, 2021 at 10:09

3 Answers 3


This should be an exercise for high school students who just learned inverse trig function.

Denote $\alpha=\tan^{-1}((2x+1)/(x+1)), \beta=\tan^{-1}((2x-1)/(x-1))$ and $\gamma=\tan^{-1}(1+x)$. The original equation is \begin{equation} (1)\qquad \alpha+\beta=2\gamma. \end{equation} The equation you solved is equivalent to \begin{equation} (2)\qquad \tan(\alpha+\beta)=\tan(2\gamma). \end{equation}

Equations $(1)$ and $(2)$ are clearly not equivalent. Actually, equation $(1)$ is equivalent to equation $(2)$ plus \begin{equation} (3)\qquad \alpha+\beta \textrm{ and } 2\gamma \textrm{ are in the same interval } ((k-\frac{1}{2})\pi, (k+\frac{1}{2})\pi), \textrm{ for some integer } k. \end{equation} It suffices to check whether your solutions of $(2)$ satisfy $(3)$ or not, which is just an easy exercise. For example, when $x=-\sqrt{2}$, then $\tan(\alpha)=1+\frac{x}{1+x}>1$, thus $\alpha=\tan^{-1}((2x+1)/(x+1))\in (\pi/4, \pi/2)$. Similarly, $\beta\in (\pi/4, \pi/2)$ and thus $\alpha+\beta\in (\pi/2,\pi)$. While $-1<1+x<0$, thus $\gamma=\tan^{-1}(1+x)\in (-\pi/4,0)$ and thus $2\gamma\in (-\pi/2,0)$. Thus $\alpha+\beta$ cannot be $2\gamma$. Instead, we should have $\alpha+\beta=2\gamma+\pi$ since $\tan(\alpha+\beta)=\tan(2\gamma)$ according your solution. The rest solutions can be checked easily. Among your solutions of $(2)$, only $x=-\sqrt{2}$ does not satisfy $(3)$ and hence does not satisfy $(1)$. All of the rest solutions satisfy (3) and thus (1), assuming your solutions of $(2)$ are correct.


I am extremely sorry for the previous answer. I had made a terrible arithmetic error, which is why I deleted the post. It seems that the roots you have are correct. If you want to see that they precisely satisfy the given equation, we do the following.

First, see that for $x = 0$, we have $LHS = \tan^{-1} 1 + \tan^{-1} 1 = 2 \tan^{-1} = RHS$.

For $x = - \dfrac{1}{2}$, again, look at the LHS.

$$\tan^{-1} \left(\dfrac{2 \left( - \frac{1}{2} \right) + 1}{- \frac{1}{2} + 1} \right) + \tan^{-1} \left( \dfrac{2 \left( - \frac{1}{2} \right) - 1}{- \frac{1}{2} - 1} \right) = \tan^{-1} \left( 0 \right) + \tan^{-1} \left( \dfrac{4}{3} \right) = \tan^{-1} \left( \dfrac{4}{3} \right).$$

Now, let $RHS = 2 \tan^{-1} \left( 1 + \left( - \frac{1}{2} \right) \right) = 2 \tan^{-1} \left( \dfrac{1}{2} \right) = \theta$. Then, we have

$$\tan \theta = \tan \left( \tan^{-1} \left( \frac{1}{2} \right) \right) = \dfrac{2 \tan \left( \tan^{-1} \left( \frac{1}{2} \right) \right)}{1 - \tan^2 \left( \tan^{-1} \left( \frac{1}{2} \right) \right)} = \dfrac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \dfrac{4}{3}.$$

Hence, $\theta = 2 \tan^{-1} \left( 1 - \frac{1}{2} \right) = \tan^{-1} \left( \frac{4}{3} \right)$. That is, $x = - \frac{1}{2}$ satisfies the equation.

Now, for $x = \sqrt{2}$, we again check $LHS$. This time, because $\left( \dfrac{2\sqrt{2} + 1}{\sqrt{2} + 1} \right) \left( \dfrac{2\sqrt{2} - 1}{\sqrt{2} - 1} \right) = 7 > 1$, we have

\begin{align*} \tan^{-1} \left( \dfrac{2\sqrt{2} + 1}{\sqrt{2} + 1} \right) + \tan^{-1} \left( \dfrac{2 \sqrt{2} - 1}{\sqrt{2} - 1} \right) &= \tan^{-1} \left( \dfrac{\frac{2\sqrt{2} + 1}{\sqrt{2} + 1} + \frac{2\sqrt{2} - 1}{\sqrt{2} - 1}}{1 - \left( \frac{2 \sqrt{2} + 1}{\sqrt{2} + 1} \right) \left( \frac{2 \sqrt{2} - 1}{\sqrt{2} - 1} \right)} \right) \\ &= \tan^{-1} \left( \dfrac{\left( 2 \sqrt{2} + 1 \right) \left( \sqrt{2} - 1 \right) + \left( 2 \sqrt{2} - 1 \right) \left( \sqrt{2} + 1 \right)}{1 - 7} \right) \\ &= \tan^{-1} \left( \dfrac{6}{-6} \right) = - \dfrac{\pi}{4}. \end{align*}

On the other hand, if we look at RHS, let us consider $2 \tan^{-1} \left( 1 + \sqrt{2} \right) = \theta$. Then,

$$\tan \theta = \dfrac{2 \left( 1 + \sqrt{2} \right)}{1 - \left( 1 - \sqrt{2} \right)^2} = \dfrac{2 + 2 \sqrt{2}}{-2 \sqrt{2} - 2} = -1.$$

Hence, $\theta = \tan^{-1} \left( -1 \right) = - \dfrac{\pi}{4}$. This tells us that $x = \sqrt{2}$ also satisfies the given condition.

Finally, for $x = - \sqrt{2}$, we consider the following:

$$\tan^{-1} \left( \dfrac{-2\sqrt{2} + 1}{-\sqrt{2} + 1} \right) + \tan^{-1} \left( \dfrac{-2\sqrt{2} - 1}{-\sqrt{2} - 1} \right) = \tan^{-1} \left( \dfrac{2 \sqrt{2} - 1}{\sqrt{2} - 1} \right) + \tan^{-1} \left( \dfrac{2 \sqrt{2} + 1}{\sqrt{2} + 1} \right) = - \dfrac{\pi}{4}.$$

On the right hand side, we consider $2 \tan^{-1} \left( 1 - \sqrt{2} \right) = \theta$. Then, we get

$$\tan \theta = \dfrac{2 \left( 1 - \sqrt{2} \right)}{1 - \left( 1 - \sqrt{2} \right)^2} = \dfrac{2 - 2 \sqrt{2}}{2\sqrt{2} - 2} = -1.$$

Hence, again, we have $\theta = 2 \tan^{-1} \left( 1 - \sqrt{2} \right) = - \dfrac{\pi}{4}$.

This gives us that all your roots satisfy the equation.

PS: While using the "subtraction identity", you need to satisfy $xy > -1$.

  • 1
    $\begingroup$ No, $-\sqrt{2}$ doesn't satisfy the equation. And tell me, with the way that you have used for verification, won't all the roots satisfy it? It is with that approach only we found those roots, problem lies while we take tan on both sides. Se, we cannot use the same thing again and say, hey! they satisfy it!. Obviously, they would. And I am looking for a method which enables me to solve this problem within 2-3 minutes since these types are asked in competitive exams (like JEE). $\endgroup$
    – V.G
    Apr 3, 2021 at 10:23
  • $\begingroup$ Yes, I got carried away! As somebody mentioned in the comments on the question, the most appropriate way to check is by looking at the domains of all the formulae we used. $\endgroup$ Apr 4, 2021 at 4:34
  • $\begingroup$ So would you edit your answer and give the correct one? Because at it's present state, it is not a correct answer. I will remove my downvote once you give a complete answer. $\endgroup$
    – V.G
    Apr 4, 2021 at 9:14

Too large for comment.

Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$\dfrac{(2x+1)(2x-1)}{(x+1)(x-1)}$ will be $\le1$


$\iff x^2-1\le0\iff-1\le x\le1$

Similarly, $(x+1)^2$ will be $\le1\iff-2\le x\le0$

For $x=-\sqrt2<-1,$ $$\tan^{-1}\dfrac{2x+1}{x+1}+\tan^{-1}\dfrac{2x-1}{x-1}=\pi+\tan^{-1}\dfrac{\dfrac{2x+1}{x+1}+\dfrac{2x-1}{x-1}}{1-\dfrac{2x+1}{x+1}\cdot\dfrac{2x-1}{x-1}}>\dfrac\pi2$$

But $$2\tan^{-1}(x+1)=2\tan^{-1}(-\sqrt2+1)=-2\tan^{-1}(\sqrt2-1)<0$$

Actually $\sqrt2-1=\csc\dfrac\pi4-\cot\dfrac\pi4=\tan\dfrac\pi8\implies\tan^{-1}(\sqrt2-1)=\dfrac\pi8$


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