tan(a+b)=3/4,tan(a-π/4)=1/2,那么tan(b+π/4)等于多少

问题描述:

tan(a+b)=3/4,tan(a-π/4)=1/2,那么tan(b+π/4)等于多少

tan(a+b)=[tan(a-π/4)+tan(b+π/4)]/[1-tan(a-π/4)tan(b+π/4)]
设tan(b+π/4)=x 3/4=(1/2+x)/(1-1/2x);x=2/11
tan(b+π/4)=2/11

2/7

b+π/4=a+b-(a-π/4)
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)}
=(3/4-1/2)/(1+3/4*1/2)
=2/11