Answer
$1$
Work Step by Step
Recall that:
$\tan(\alpha+\beta)=\dfrac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$.
Hence,
$\dfrac{\tan(20^\circ)+\tan(25^\circ)}{1-\tan(20^\circ)\tan(25^\circ)}=\tan(20^\circ+25^\circ)=\tan(45^\circ)=1$
