∵2A=B+C,A+B+C=2A+A=180°
∴A=60 B>0,C>0,B+C=120°,设 C=120°-B=120°-x
则,tanB+tanC=sin(B+C)/cosBcosC.
=sin(x+120°-x)/cosx*cos(120°-x).
=sin120°/{(1/2)[cos(x+120°-x)+cos(x-(120°-x))]} .
=√3/[cos120°+cos(2x-120°)
=√3/{[(-1/2)+sin[90°+(2x-120°)]|.
tanB+tanC=√3/[sin(2x-30°)-(1/2)] [ 0＜(2x-30°)≤120°]
0＜sin(2x-30≤ 1.
∴ -2√3 ≤tanB+tanC≤2√3.
√√