求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3)
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求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3)求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cos
求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3)
求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3)
求证（tanαtan2α/tan2α-tanα）+√3[(sinα)^2-(cosα)^2]=2sin(2α-π/3)
[tanαtan2α/(tan2α-tanα）]+√3[(sinα)^2-(cosα)^2]
=sinasin2a/cosacos2a)/(sin2a/cos2a-sina/cosa)-√3cos2a
=sinasin2a/cosacos2a)/[(sin2acosa-cos2asina)/cos2acosa]-√3cos2a
=sinasin2a/sina-√3cos2a
=2x(1/2)sin2a-2x(√3/2)cos2a
=2sin(2a-π/3)
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