题目应该是：“sin(α+β)=1,求证：tan(2α+β)+tanβ=0 ”吧证明：sin(a+b)=1 →cos(a+b)=√[1-sin^2(a+b)]=0 →sin(2a+2b)=2*sin(a+b)*cos(a+b)=0 →tan(2a+2b)=sin(2a+2b)/cos(2a+2b)=0 tan(2a+b)+tanb=tan(2a+2b...