Answer
$0$
Work Step by Step
Since $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}}$
Then,
$\dfrac{\sin{40^{\circ}}}{\cos{40^{\circ}}} = \tan{40^{\circ}}$
Therefore,
$\tan{40^{\circ}} - \dfrac{\sin{40^{\circ}}}{\cos{40^{\circ}}} = \tan{40^{\circ}}-\tan{40^{\circ}} = \boxed{0}$
