已知tan(a+b)=1/2,tan(a-π/4)= -1/3,求tan(b+π/4)的值

问题描述:

已知tan(a+b)=1/2,tan(a-π/4)= -1/3,求tan(b+π/4)的值

tan(b+π/4)=tan[(a+b)-(a-π/4)]=[tan(a+b)-tan(a-π/4)]/[1+tan(a+b)tan(a-π/4)]=[1/2-(-1/3)]/[1+1/2*(-1/3)]=1