Prove that: $\;$ $\cot^{-1} \left(\dfrac{\sqrt{1 + x^2} -1}{x}\right) = \dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} \left(x\right)$
Let $x = \tan \left(\theta\right) \implies \theta = \tan^{-1} \left(x\right), \hspace{1cm} \theta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
$\begin{aligned}
\therefore \; \cot^{-1} \left(\dfrac{\sqrt{1 + x^2} -1}{x}\right) & = \cot^{-1} \left(\dfrac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta}\right) \\\\
& = \cot^{-1} \left(\dfrac{\sec \theta - 1}{\tan \theta}\right) \\\\
& = \cot^{-1} \left(\dfrac{1 - \cos \theta}{\sin \theta}\right) \\\\
& = \cot^{-1} \left[\dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)}\right] \\\\
& = \cot^{-1} \left[\tan \left(\dfrac{\theta}{2}\right)\right] \\\\
& = \cot^{-1} \left[\cot \left(\dfrac{\pi}{2} - \dfrac{\theta}{2}\right)\right] \\\\
& = \dfrac{\pi}{2} - \dfrac{\theta}{2} \\\\
& = \dfrac{\pi}{2} - \dfrac{1}{2} \tan^{-1} \left(x\right)
\end{aligned}$
Hence proved.