Use tan⁻¹(√(1 + x²) - 1)/x = 1/2 tan⁻¹x to find the value of the left side at x = √3 and at x = -√3…

Step 1. At $x = \sqrt{3}$. The result is $\dfrac{1}{2}\tan^{-1}\sqrt{3} = \dfrac{1}{2} \times \dfrac{\pi}{3} = \dfrac{\pi}{6} = 0.5236$.

Step 2. A direct check. $\sqrt{1 + 3} = 2$, so the fraction is $\dfrac{2 - 1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}}$, and $\tan^{-1}\dfrac{1}{\sqrt{3}} = \dfrac{\pi}{6}$.

Step 3. At $x = -\sqrt{3}$. The fraction is $\dfrac{2 - 1}{-\sqrt{3}} = -\dfrac{1}{\sqrt{3}}$, and the value is $-\dfrac{\pi}{6} = -0.5236$. This agrees with the rule, because the form holds for every $x \ne 0$.