Question

已知sin(a+b)=1,求证：tan(2a+b)+tanb=0

Answer 1

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

Answer 2

sin(a+b)=1,所以：a+b=2kπ+π/2,k为正整数。
所以：
tan(2a+b)+tanb
=tan(2a+2kπ+π/2-a)+tan(2kπ+π/2-a)
=tan(a+π/2)+tan(π/2-a)
=-ctga+ctga
=0
得证。