Discussion
If \(f(x) = \tan^{-1}\left(\dfrac{2x}{1 - x^{2}}\right)\), then \(f\left(\dfrac{1}{\sqrt{3}}\right)\) is equal to
Answer: C
\(f\left(\frac{1}{\sqrt{3}}\right) = \) \(\frac{\pi}{3}\).
Substitute directly:
\[ \begin{aligned} \frac{2x}{1 - x^{2}} &= \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3} = \sqrt{3} \\ f &= \tan^{-1}\sqrt{3} = \frac{\pi}{3} \end{aligned} \]
This agrees with the identity \(\tan^{-1}\frac{2x}{1 - x^{2}} = 2\tan^{-1}x\) for \(|x| \lt 1\): \(2 \cdot \frac{\pi}{6} = \frac{\pi}{3}\). \(\frac{\pi}{6}\) forgets the factor 2; \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) lie outside the principal range \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
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