Solutions to inverse trig equation $\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi}{3}$

Question

I have to find the solutions of the equation,

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi}{3}$$

So I know that \begin{align}\tan^{-1}\left(\frac{2x}{1-x^2}\right) &= 2\tan^{-1}x\\ \cot^{-1} y &=\tan^{-1}\frac{1}{y}\end{align}

Using these I found the solution to be $$x = \frac{1}{\sqrt3}$$

But my book says that $-\sqrt3$, $\sqrt3 +2$, $\sqrt3 -2$ are also solutions.

How do I obtain those?

Answer 1

Before I start solving, I would like to say that the formula you mentioned:

$$\cot^{-1}(x) =\tan^{-1}\left(\frac{1}{x}\right)$$

is not always true. You can verify this by graphing $\cot^{-1}(x)$ and $\tan^{-1}\left(\frac{1}{x}\right)$ on Desmos. This is only correct if $x > 0$.

Here is the correct statement (from Wikipedia):

If $x>0$, then: $$\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$$ If $x<0$, then: $$\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) + \pi$$

So to solve the problem, let's consider 2 cases:

Case $1$: $\frac{1-x^2}{2x} > 0 \text{ (1)}$

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi}{3}$$ $$ \implies \tan^{-1}\left(\frac{2x}{1-x^2}\right)+\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\pi}{3}$$ $$ \implies 2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\pi}{3}$$ $$ \implies \tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}$$ $$ \implies \frac{2x}{1-x^2}= \tan\left(\frac{\pi}{3}\right)$$ $$ \implies \frac{2x}{1-x^2}= \sqrt{3}$$ $$ \implies \sqrt{3} \text{ } x^2 + 2x - \sqrt{3} = 0$$ $$ \implies x = -\sqrt{3} \text{ or } x = \frac{1}{\sqrt{3}} \text{ (Both satisfy (1)) }$$

Case $2$: $\frac{1-x^2}{2x} < 0 \text{ (2) }$

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi}{3}$$ $$\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right) - \pi =\frac{2\pi}{3} - \pi$$ $$ \implies \tan^{-1}\left(\frac{2x}{1-x^2}\right)+\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{-\pi}{3}$$ $$ \implies 2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{-\pi}{3}$$ $$ \implies \tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{-\pi}{6}$$ $$ \implies \frac{2x}{1-x^2}= \tan\left(-\frac{\pi}{6}\right)$$ $$ \implies \frac{2x}{1-x^2}= -\frac{1}{\sqrt{3}}$$ $$ \implies x^2 - 2\sqrt{3} \text{ }x - 1 = 0$$ $$ \implies x = \sqrt{3} \pm 2 \text{ (Both satisfy (2)) }$$

Answer 2

We have \begin{align}\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)&=\frac{2\pi}{3}\tag1\\ \implies 2\tan^{-1}(x)-2\cot^{-1}(x)&=\frac{2\pi}{3}\\ \implies \tan^{-1}(x)-\cot^{-1}(x)&=\frac{\pi}{3}\\ \implies \tan^{-1}(x)-\tan^{-1}\left( \dfrac{1}{x} \right)&=\frac{\pi}{3}\\ \implies \tan^{-1}\left( \dfrac{x-1/x}{1+1} \right)&=\frac{\pi}{3}\\ \implies \dfrac{x-1/x}{2} &=\tan\frac{\pi}{3}\\ \implies \dfrac{x-1/x}{2} &=\sqrt 3\\ \implies x^2-2\sqrt 3x-1&=0\\ \implies x&=\dfrac{2\sqrt 3\pm\sqrt{12+4}}{2}\\ \implies x&=\sqrt 3\pm 2\end{align}

We know $$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left( \dfrac{x+y}{1−xy} \right)~,$$ $$\cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1}\left( \dfrac{xy- 1}{y + x} \right)$$

Second method :

Put $~x=\tan \theta~$, equation $(1)$ becomes

\begin{align}\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\left(\frac{1-x^2}{2x}\right)&=\frac{2\pi}{3}\\ \implies \tan^{-1}\left(\frac{2\tan \theta}{1-\tan^2 \theta}\right)+\cot^{-1}\left(\frac{1-\tan^2 \theta}{2\tan \theta}\right)&=\frac{2\pi}{3}\\ \implies \tan^{-1}\left(\tan2\theta\right)+\cot^{-1}\left(\cot2\theta\right)&=\frac{2\pi}{3}\\ \implies 2\theta+2\theta=\frac{2\pi}{3}\\ \implies \theta=\frac{\pi}{6}\\ \implies \tan^{-1}x=\frac{\pi}{6}\\ \implies x=\tan\left(\frac{\pi}{6}\right)=\dfrac{1}{\sqrt 3} \end{align}

Answer 3

The answer depends on the range specification of the inverse trigonometric function $\cot^{-1}t$, which is not yet universally agreed upon. There are two common choices,

$$\cot^{-1} t \in (0,\pi),\>\>\>\>\> \text{or}\>\>\>\>\> \cot^{-1} t \in [-\frac\pi2,0) \cup(0, \frac\pi2]$$

The first specification is more traditional, while the second is also widely accepted (for example, endorsed by WolfAlpha).

So, depending on the range choices for the function $\cot^{-1}t$, you would get different answers to the problem. Shown below in the graph are the functional values of the two choices for the given function and the corresponding solutions. The first choice produces the four roots given in your book and the second choice would just give $-\sqrt3$ and $\frac1{\sqrt3}$ as roots.

Your solution, albeit incomplete, corresponds to the second choice that permits $\cot^{-1}\frac1t = \tan^{-1}t$. Then, you have $\tan^{-1}\frac{2x}{1-x^2}=\frac{\pi}{3}$, or $\frac{2x}{1-x^2}=\sqrt3$ with $-\sqrt3$ and $\frac1{\sqrt3}$ as roots.