若tana=√3 ,则cos2a-cos^2 a / sin^2 a-1=

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若tana=√3,则cos2a-cos^2a/sin^2a-1=若tana=√3,则cos2a-cos^2a/sin^2a-1=若tana=√3,则cos2a-cos^2a/sin^2a-1=sin&

若tana=√3 ,则cos2a-cos^2 a / sin^2 a-1=
若tana=√3 ,则cos2a-cos^2 a / sin^2 a-1=

若tana=√3 ,则cos2a-cos^2 a / sin^2 a-1=
sin²a+cos²=1
所以sin²a-1=-cos²a
原式=(cos²a-sin²a-cos²a)/(sin²a-1)
=-sin²a/(-cos²a)
=tan²a
=3

由tana=√3,有sin^2a/cos^2a=3,sin^2a=3cos^2a,4cos^2a=1,cos^2a=1/4
于是,原式=2cos^2a-1-sin^2a/cos^2a-1=1/2-2-1/3=-7/6。