Miscellaneous · Q12: Solve $\tan^{-1}\frac{1-x}{1+x}=\frac12\tan^{-1}x$

Basics you need

Solution

1. Name the angle and locate it

   Let $\varphi = \tan^{-1}x$⁠, so $\tan\varphi = x$⁠. Since $x > 0$⁠,

   $$0 < \varphi < \frac\pi2.$$
   Writing the equation in terms of the single angle $\varphi$ is the aim. The interval for $\varphi$ comes from the principal range of $\tan^{-1}$ together with $x > 0$ (a positive tan value comes from an angle in $\left(0,\frac\pi2\right)$⁠).
   Principal range of $\tan^{-1}$ (Basic 1). Lesson: substitutions and simplest form

2. The left side is π/4 − φ

   Using Basic 2,

   $$\frac{1-x}{1+x} = \frac{1-\tan\varphi}{1+\tan\varphi} = \tan\left(\frac\pi4-\varphi\right).$$

   From $0<\varphi<\frac\pi2$ we get $-\frac\pi4 < \frac\pi4-\varphi < \frac\pi4$⁠, which is inside $\left(-\frac\pi2,\frac\pi2\right)$⁠. So

   $$\tan^{-1}\frac{1-x}{1+x} = \frac\pi4 - \varphi.$$
   $\tan^{-1}(\tan\alpha)$ returns $\alpha$ only if $\alpha$ is in the principal range. We checked that $\frac\pi4-\varphi$ is, so the rewrite is exact for every $x > 0$⁠. Also $1 + x > 0$⁠, so the fraction is always defined.
   $\tan^{-1}(\tan\alpha) = \alpha$ for $\alpha\in\left(-\frac\pi2,\frac\pi2\right)$ (Basic 1). Lesson: the folding rules

3. Solve the equation for φ, then x

   The equation becomes

   $$\frac\pi4 - \varphi = \frac12\varphi \iff \frac32\varphi = \frac\pi4 \iff \varphi = \frac\pi6.$$

   This $\varphi$ is in $\left(0,\frac\pi2\right)$⁠, as required. So

   $$x = \tan\frac\pi6 = \frac{1}{\sqrt3},$$

   which satisfies $x > 0$⁠.

   Each step here is reversible (adding the same quantity to both sides, multiplying by a non-zero number), and for $\varphi\in\left(0,\frac\pi2\right)$ the value of $\varphi$ determines $x = \tan\varphi$ uniquely. So $x = \frac1{\sqrt3}$ is the only candidate.
   $\tan\frac\pi6 = \frac1{\sqrt3}$ (Basic 3).

4. Check in the original equation

   With $x = \frac{1}{\sqrt3}$⁠, multiply top and bottom of the fraction by $\sqrt3$⁠:

   $$\begin{aligned} \frac{1-x}{1+x} &= \frac{\sqrt3-1}{\sqrt3+1} = \frac{(\sqrt3-1)^2}{3-1} \\ &= \frac{4-2\sqrt3}{2} = 2-\sqrt3. \end{aligned}$$

   So the left side is $\tan^{-1}(2-\sqrt3) = \frac\pi{12}$⁠, and the right side is $\frac12\tan^{-1}\frac{1}{\sqrt3} = \frac12\cdot\frac\pi6 = \frac\pi{12}$ ✓.

   A check in the original equation confirms the algebra and the principal-range reasoning together. Multiplying by $\sqrt3-1$ top and bottom (rationalising) is how $2-\sqrt3$ appears.
   $\tan\frac\pi{12} = 2-\sqrt3$⁠, and $\frac\pi{12}$ is in the principal range (Basic 3).

5. No other solution, and a trap to avoid

   Uniqueness. For $x > 0$ the left side $\frac\pi4 - \tan^{-1}x$ strictly decreases, and the right side $\frac12\tan^{-1}x$ strictly increases. So their difference (left minus right) strictly decreases and can be zero at most once, so $\frac1{\sqrt3}$ is the only solution.

   Another route, and its false root. Doubling both sides gives $2\tan^{-1}t = \tan^{-1}x$ with $t = \frac{1-x}{1+x}$⁠. For $x > 0$⁠, $|t| < 1$⁠, so Basic 4 applies. Since $(1+x)^2-(1-x)^2 = 4x$⁠,

   $$\frac{2t}{1-t^2} = \frac{2(1-x)(1+x)}{4x} = \frac{1-x^2}{2x}.$$

   Setting this equal to $x$⁠: $1 - x^2 = 2x^2$⁠, so $x^2 = \frac13$ and $x = \pm\frac{1}{\sqrt3}$⁠.

   The root $-\frac1{\sqrt3}$ is rejected by $x > 0$⁠. It is also not a solution of the original equation: there $t = 2+\sqrt3 > 1$⁠, the left side is $\tan^{-1}(2+\sqrt3) = \frac{5\pi}{12}$⁠, but the right side is $-\frac{\pi}{12}$⁠.

   The false root appears because $1-x^2 = 2x^2$ only remembers that the two sides have equal tangents; for $x<0$ we have $|t|>1$⁠, the double-angle rule fails, and equal tangents no longer mean equal angles. The method of Steps 1 to 3 never leaves the principal range, so it cannot produce such a root.
   $2\tan^{-1}t = \tan^{-1}\frac{2t}{1-t^2}$ only for $|t|<1$ (Basic 4).

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