Evaluate: $\;$ $\sin \left[2 \tan^{-1} \left(\dfrac{2}{5}\right)\right]$
Let $\tan^{-1} \left(\dfrac{2}{5}\right) = \theta \;\;\; \text{ where } \theta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
Then, $\tan \left(\theta\right) = \dfrac{2}{5}$
$\begin{aligned}
\therefore \; \sin \left[2 \tan^{-1} \left(\dfrac{2}{5}\right)\right] & = \sin \left(2 \theta\right) \\\\
& = \dfrac{2 \tan \theta}{1 + \tan^2 \theta} \\\\
& = \dfrac{2 \times \dfrac{2}{5}}{1 + \dfrac{4}{25}} \\\\
& = \dfrac{4 / 5}{29 / 25} \\\\
& = \dfrac{20}{29}
\end{aligned}$