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I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$

But I missed a solution (don't know where's the mistake in my work).

Here's my work: $$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$ Putting $x = \tan(\theta)$ $$\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\ \tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\ \frac\pi 4 - \theta&=\frac{1}{2}\theta\\ \frac\pi 4&=\frac{1}{2}\theta + \theta\\ \frac\pi 4&=\frac{3}{2}\theta\\ \frac\pi 6&=\theta\\ \frac\pi 6&=\tan^{-1}(x)\\ \tan\frac\pi 6&=(x)\\ \frac1{\sqrt{3}}&=x\end{align}$$

I got only one solution, but the answer in my textbook is $\pm\frac{1}{\sqrt{3}}$. Where is the mistake?

3 Answers3

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$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$ Putting $x = \tan(\theta)$

Here, it is useful to impose the restriction $$\theta\in\left(-\frac\pi2,-\frac\pi4\right)\cup\left(-\frac\pi4,\frac\pi2\right);\tag1$$ since $\tan\theta$ for which the given equation is defined is surjective on this interval, this restriction is valid and does not contract the solution set.

\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\ \tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\ \frac\pi 4 - \theta&=\frac{1}{2}\theta\end{align}

Note that $$\arctan(\tan \alpha)\not\equiv \alpha;\tag{*}$$ instead, for each $\alpha\in\mathbb R,$ there is some $k\in\mathbb Z$ such that $$\arctan(\tan \alpha)=\alpha+k\pi.$$ By condition $(1),$ $$\arctan \left(\tan\theta\right)=\theta$$ while \begin{align}\arctan\left(\tan\left(\frac\pi4 -\theta\right)\right)&=\frac\pi4 -\theta \quad\text{or}\quad \left(\frac\pi4 -\theta\right)-\pi\\ &=\frac\pi4 -\theta \quad\text{or}\quad -\frac34\pi -\theta.\end{align}

\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\end{align}

Continuing with your substitution method: \begin{align} \frac12\arctan \left(\tan\theta\right) &=\arctan\left(\tan\left(\frac\pi4 -\theta\right)\right)\\ \frac12\big(\theta\big) &= \left(\frac\pi4 -\theta\right) \quad\text{or}\quad \left(-\frac34\pi -\theta\right)\\ \theta &= \frac\pi6 \quad\text{or}\quad -\frac\pi2\\ &= \frac\pi6\\ x&=\frac1{\sqrt3}.\end{align} Plugging this value into the given equation reveals that it is the only solution. So, while you had obtained the correct solution set by serendipity (without proper justification), your textbook is actually wrong too in neglecting to eliminate its extraneous solution $x=-\dfrac1{\sqrt3}.$

ryang
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    This is the only correct answer since it mentions $\arctan({\tan\alpha})\neq \alpha$ – Lalit Tolani May 17 '22 at 09:17
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    Thank you, that's totally different from what I've been taught in school. –  May 17 '22 at 09:26
  • My solution is not properly justified because I didn't restrict the value of $\theta$ to $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Right? –  May 17 '22 at 15:43
  • @Yooo No: restricting $\theta$ to $\arctan$'s principal range makes your substitution $x=\tan\theta$ injective, which merely makes it subsequently easier to recover $x,$ but this step is not crucial. Your mistake is in using the false identity $\arctan(\tan \alpha)\equiv \alpha.$ – ryang May 17 '22 at 15:54
  • Ohhh, actually in my textbook, they are given as identities that, $\arcsin(\sin(x)) = x, \arccos(\cos(x)) = x ...$ for the 6 trig. functions. –  May 17 '22 at 15:57
  • @Yooo Since $\arccos(\cos(2\pi))=0,$ we also know that $\arccos(\cos(x))$ does not generally equal $x$ either. Similarly for $\sin.$ The equalities do hold though for all $x$ in the principal range. – ryang May 17 '22 at 16:01
  • Yes. The point is that these identities work well only when $x$ lies in the principal branch which the book doesn't mention. –  May 17 '22 at 16:07
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    Yes exactly! I think I should buy a better book lol –  May 17 '22 at 16:08
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Apply $tan$ to both sides

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

From this,

$ (1 - x) (\sqrt{1 + x^2} + 1 ) = x (1 + x) $

$ (1 - x) \sqrt{1 + x^2} + 1 - x = x + x^2 $

$ \sqrt{ 1 + x^2} = \dfrac{(x^2 + 2 x - 1)}{ ( 1 - x)} $

$ 1 + x^2 = \dfrac{ (x^2 + 2x - 1)^2 }{ (x - 1)^2 } $

$ (x^2 + 1) (x - 1)^2 = (x^2 + 2 x - 1)^2 $

$ (x^2 + 1)(x^2 - 2 x + 1) = x^4 + 4 x^2 + 1 + 4 x^3 - 4 x - 2 x^2 $

$ x^4 - 2 x^3 + x^2 + x^2 - 2 x + 1 = x^4 + 4 x^3 + 2 x^2 - 4 x + 1 $

Cancelling $x^4 $ and $1$ on both sides, and re-arranging,

$ 6 x^3 - 2 x = 0 $

Dividing by $(2 x)$ (The root $0$ is extraneous)

$ 3 x^2 - 1 = 0 $

Hence, $ x = \pm \dfrac{1}{\sqrt{3}} $

Hosam Hajeer
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  • How did you get $\frac{ x }{ \sqrt{1 + x^2} + 1 }$ from $\tan(\frac12\tan^{-1}(x))$? May you explain please? –  May 15 '22 at 03:12
  • $\tan(\frac{1}{2} \phi) = \dfrac{ \sin(\frac{1}{2} \phi )}{\cos(\frac{1}{2} \phi) } = \dfrac{ \cos(\frac{1}{2} \phi) \sin(\frac{1}{2} \phi)}{ \cos^2(\frac{1}{2} \phi)} = \dfrac{ \sin(\phi) }{1 + \cos(\phi) } \ = \dfrac{ \tan(\phi) }{ \sec(\phi) + 1 } \ = \dfrac{ \tan(\phi) }{ \sqrt{\tan^2(\phi) + 1} + 1 } $ – Hosam Hajeer May 15 '22 at 10:40
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Thank you to @Robin'sPremiumCoffee for a nice algebraic solution. Now the problem with your original method is that although: $$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)$$ It is also equal to: $$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)=\tan(\pi +\frac \pi 4 -\theta)$$ This is because $\tan(\theta)$ is $\pi$ periodic. From there we may continue your steps to get: \begin{align} \pi + \frac\pi 4 - \theta&=\frac{1}{2}\theta\\ \frac{5\pi} 4&=\frac{1}{2}\theta + \theta\\ \frac{5\pi} 4&=\frac{3}{2}\theta\\ \frac{5\pi} 6&=\theta\\ \frac{5\pi}6&=\tan^{-1}(x)\\ \tan\frac{5\pi} 6&=(x)\\ -\frac1{\sqrt{3}}&=x\end{align}

person
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