Question

已知tana,tanb是方程7x^2-8x+1=0的两个根,求tan(a-b)

速度

Answer 1

由韦达定理知：tana+tanb=8/7，tanatanb=1/7
(tana-tanb)^2=(tana+tanb)^2-4tanatanb=(8/7)^2-4*(1/7)=36/49
所以tana-tanb=±6/7
tan(a-b)=(ana-tanb)/(1-tanatanb)=(±6/7)/(1-1/7)=±1

Answer 2

方程的两根为1/7,1,即tana=1/7,tanb=1.tan(a-b) =（tana-tanb）/（1+tana·tanb）=3/4或-3/4

Answer 3

tana+tanb=8/7
tanatanb=1/7
(tana-tanb)^2=(tana+tanb)-4tanatanb
=(8/7)^2-4/7
=64/49-28/49
=36/49
tana-tanb=±6/7
tan(a-b)=(tana-tanb)/(1+tanatanb)
=±6/7/(1+1/7)
=±6/(7+1)
=±6/8
=±3/4