若0
问题描述:
若0
答
tan(x+y)=(tany+tanx)/(1-tanx*tany)
tan(x+y)=1
and 0
答
tanx=1/7,tany=3/4
则tan(x+y)=(tanx+tany)/(1-tanxtany)=(1/7+3/4)/[1-(1/7)*(3/4)]=1
又0<x<π/2,0<y<π/2
则0<x+y<π
所以x+y=π/4
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