Answer:
[tex]\displaystyle x = \frac{1}{3}[/tex].
Step-by-step explanation:
Consider the double angle identity for tangents:
[tex]\displaystyle \tan(2\theta) = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}[/tex].
Also, for two complementary angles,
[tex]\displaystyle \tan{\left(\frac{\pi}{2} - \theta \right)} = \frac{1}{\tan{\theta}}[/tex].
Subtract [tex]\tan^{-1}(x + 1)[/tex] from both sides of this equation:
[tex]\displaystyle 2\tan^{-1}{x} = \frac{\pi}{2} + (-\tan^{-1}(x + 1))[/tex].
Take the tangent of both sides of this equation:
[tex]\displaystyle \tan(2\tan^{-1}{x}) = \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)[/tex].
Apply the double-angle identity to the left-hand side of this equation:
[tex]\displaystyle \tan(2\tan^{-1}{x}) \implies \frac{2\tan(\tan^{-1}x)}{1 - \tan^{2}(\tan^{-1}x)}\implies \frac{2x}{1 - x^{2}}[/tex].
The two angles [tex]\displaystyle \left(\frac{\pi}{2} + (-\tan^{-1}\left(x + 1\right))\right)[/tex] and [tex](x - 1)[/tex] are complementary. Therefore, for the right-hand side of this equation,
[tex]\displaystyle \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)\implies \frac{1}{\tan(\tan^{-1}(x + 1))} \implies \frac{1}{x + 1}[/tex].
Equate the two sides of this equation:
[tex]\begin{aligned} & \frac{2x}{1 - x^{2}} = \frac{1}{x + 1}\\ \implies & \frac{2x}{(1 - x)(1 + x)} = \frac{1}{1 + x}\\\implies & 2x = 1 - x,\; x \ne 1,\; x \ne -1\\\implies & x = \frac{1}{3}\end{aligned}[/tex].
