Question

一道函数题求助

已知tan（a+b）=3，tan（a-b）=5，求tan2a的值
需要过程，谢谢

Answer 1

tan(a+b)+tan(a-b)=sin(a+b)/cos(a+b)+sin(a-b)/cos(a-b)
=[sin(a+b)cos(a-b)+sin(a-b)cos(a+b)]/(cos(a+b)cos(a-b))
=sin(a+b+a-b)/(cos(a+b)cos(a-b))=3
sin2a=3cos(a+b)cos(a-b)

tan(a+b)*tan(a-b)=[sin(a+b)sin(a-b)]/[cos(a+b)cos(a-b)]=5
sin(a+b)sin(a-b)=5cos(a+b)cos(a-b)
cos(a+b)cos(a-b)-sin(a+b)sin(a-b)=cos(a+b+a-b)
=cos2a=-4cos(a+b)cos(a-b)
tan2a=sin2a/cos2a=-3/4