已知cosa=-4/5,a(π/2,π),tan(π/4+a)等于

问题描述:

已知cosa=-4/5,a(π/2,π),tan(π/4+a)等于

a(π/2,π),
所以sina>0
sin²a+cos²a=1
所以sina=3/5
tana=sina/cosa=-3/4
tanπ/4=1
所以原式=(tana+tanπ/4)/(1-tanatanπ/4)=1/7

cosa=-4/5
sina=3/5
tana=-3/4
tan(π/4+a)
=(tanπ/4+tana)/(1-tanπ/4tana)
=(1+tana)/(1-tana)
=1/7