tan(兀/4-a)=3,则cos2a/1+sin2a=
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tan(兀/4-a)=3,则cos2a/1+sin2a=tan(兀/4-a)=3,则cos2a/1+sin2a=tan(兀/4-a)=3,则cos2a/1+sin2a=tana=tan(π/4-(兀/
tan(兀/4-a)=3,则cos2a/1+sin2a=
tan(兀/4-a)=3,则cos2a/1+sin2a=
tan(兀/4-a)=3,则cos2a/1+sin2a=
tana=tan(π/4-(兀/4-a)=(1-3)/(1+3)=-1/2
cos2a/1+sin2a=(cos²a-sin²a)/(sina+cosa)²=(sina-cosa)/(sina+cosa)
=(1-tana)/(1+tana)
=3
不懂再问,不客气!
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