$解：??(1)tan A=\frac {a}b=\frac {15}{9}=\frac {5}{3}，????tan B=\frac b{a}=\frac {9}{15}=\frac {3}{5}??$
$??(2)??不妨令??a=2，??則??c=3??$
$由勾股定理可得??b=\sqrt {c^2-a^2}=\sqrt 5??$
$∴??tan A=\frac {a}b=\frac 2{\sqrt 5}=\frac {2\sqrt 5}5，????tan B=\frac b{a}=\frac {\sqrt 5}2??$
$??(3)??不妨令??b=1，??則??c=4??$
$由勾股定理可得??a=\sqrt {c^2-b^2}=\sqrt {15}??$
$∴??tan A=\frac {a}b=\sqrt {15}，????tan B=\frac b{a}=\frac 1{\sqrt {15}}=\frac {\sqrt {15}}{15}?$
