Question

已知sin(a+b)=1,求证tan(2a+b)+tanb=0,

Answer 1

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-b)+tanb

=[tan(2a+2b)-tanb]/[1+tan(2a+2b)tanb]+tanb

=[0-tanb]/[1+0*tanb]+tanb

=-tanb+tanb

=0