已知sin(a+b)=1,求证：tan(2a+b)+tanb=0
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已知sin(a+b)=1,求证：tan(2a+b)+tanb=0已知sin(a+b)=1,求证：tan(2a+b)+tanb=0已知sin(a+b)=1,求证：tan(2a+b)+tanb=0sin(
已知sin(a+b)=1,求证：tan(2a+b)+tanb=0
已知sin(a+b)=1,求证：tan(2a+b)+tanb=0
已知sin(a+b)=1,求证：tan(2a+b)+tanb=0
sin(a+b)=1,cos(a+b)=0
tan(2a+b)+tanb=tan[a+(a+b)]+tanb
=[sinacos(a+b)+cosasin(a+b)]/[cosacos(a+b)-sinasin(a+b)]+tanb
=-cosa/sina+tanb=(sinbsina-cosacosb)/sinacosb=-cos(a+b)/sinacosb=0
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