做一题:已知x+y=pie/3,且√3(tanxtany+c)+tanx=0(c为常数),求tanx的值?(pie原为3.1415...,这里是弧度制)

Question

做一题:已知x+y=pie/3,且√3(tanxtany+c)+tanx=0(c为常数),求tanx的值?(pie原为3.1415...,这里是弧度制)

Answer 1

tan(x+y)=(tanx+tany)/(1-tanxtany)
tanxtany=1-(tanx+tany)/tan(x+y)=1－√3/3(tanx+tany)
√3(tanxtany+c)+tanx=0=√3[1－√3/3(tanx+tany)+c]+tanx
整理得tany=√3(1+c)