Question

已知cos(a+b)=0,求证：tan(2a+b)+tanb=o？

Answer 1

cos(a+b)=0
sin(a+b)=1
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-cosaco *** )/sinaco ***
=-cos(a+b)/sinaco ***
=0,8,∵cos(a+b)=0，∴a+b=kπ+π/2 （式中k是整数）
tan(2a+b)=tan[2(a+b)-b]
=tan[2(kπ+π/2)-b]
=tan[2kπ+π-b]
=tan(π-b)
=-tanb,，
∴tan(2a+b)+tanb=0。,1,把cos《a+b》=0化简，带入要求证的,1,