Compute the given expression: $\;$ $\tan \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{5}{13}\right)\right]$
Given expression: $\;$ $\tan \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{5}{13}\right)\right]$
$= \dfrac{1 - \cos \left[\sin^{-1} \left(\dfrac{5}{13}\right)\right]}{\sin \left[\sin^{-1} \left(\dfrac{5}{13}\right)\right]}$ $\;\;\;$ $\left\{\because \; \tan \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{\sin \theta}\right\}$
$= \dfrac{1 - \sqrt{1 - \left(\dfrac{5}{13}\right)^2}}{\dfrac{5}{13}}$ $\;\;\;$ $\left\{\because \; \sin \left(\sin^{-1} x\right) = x \; \forall \; x \in \left[-1,1\right], \right.$
$\hspace{4.5cm} \left. \cos \left(\sin^{-1} x\right) = \sqrt{1- x^2}, \; x \geq 0 \right\}$
$= \dfrac{1 - \dfrac{12}{13}}{\dfrac{5}{13}}$
$= \dfrac{1}{5}$