Calculate: $\;$ $\sin \left[2 \tan^{-1} \left(3\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \tan^{-1} \left(3\right)\right]$
$= 2 \times \sin \left[\tan^{-1} \left(3\right)\right] \cos \left[\tan^{-1} \left(3\right)\right]$
$= 2 \times \sin \left[\sin^{-1} \left(\dfrac{3}{\sqrt{1 + 3^2}}\right)\right] \cos \left[\cos^{-1} \left(\dfrac{1}{\sqrt{1 + 3^2}}\right)\right]$
$= 2 \times \dfrac{3}{\sqrt{10}} \times \dfrac{1}{\sqrt{10}}$
$= \dfrac{3}{5}$
Formulas used:
$\sin 2 \theta = 2 \sin \theta \cos \theta$
$\tan^{-1} x = \sin^{-1} \left(\dfrac{x}{\sqrt{1 + x^2}}\right), \;\; x > 0$
$\tan^{-1} x = \cos^{-1} \left(\dfrac{1}{\sqrt{1 + x^2}}\right), \;\; x > 0$
$\sin \left(\sin^{-1} x\right) = x, \;\; \forall \; x \in \left[-1, 1\right]$
$\cos \left(\cos^{-1} x\right) = x, \;\; \forall \; x \in \left[-1, 1\right]$