Answer
$\dfrac{\sqrt3}{3}$
Work Step by Step
Recall that:
$\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$
Hence,
$\dfrac{\tan(40^\circ)-\tan(10^\circ)}{1+\tan(40^\circ)\tan(10^\circ)}\\
=\tan(40^\circ-10^\circ)\\
=\tan(30^\circ)\\
=\dfrac{\sqrt3}{3}$
