tan(2009π+a)=3
即tan(π+a)=3,tana=3
a在第一象限时,sina=3√10/10,cosa=√10/10
a在第三象限时,sina=-3√10/10,cosa=-√10/10
sin[(a-3π)-2009π/2+a]/sin(-a)+cos(π+a)
=-sin(2a-π-π/2)/sina-cosa
=sin[π-(2a-π/2)]/sina-cosa
=-sin(π/2-2a)/sina-cosa
=-cos2a/sina-cosa
=-cosa/tana+sina-cosa
(1)=-√10/30+3√10/10-√10/10
=√10/6
(2)=-√10/6