解法一:(1-√3tan75°)/(√3+tan75°)=(1-tan60°tan75°)/(tan60°+tan75°)=1/tan(60°+75°)=1/tan135°=-1解法二:tan75°=tan(45°+30°)=(tan45°+tan30°)/(1-tan45°tan30°)=(1+√3/3)/(1-√3/3)...
tanx=1/3,tan(x-y)=-2,且π/2<y<x,求y
答:
tanx=1/3,tan(x-y)=-2
所以:
tan(x-y)=(tanx-tany)/(1-tanxtany)=-2
所以:
(1/3-tany)/(1-tany/3)=-2
1/3-tany=-2+(2/3)tany
tany=7/3