怎么做？？？？？？？？？ b）题
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2(sin2θ-cos2θ+1)(sin2θ+cos2θ+1)/[(sin2θ+cos2θ+1)²-(sin2θ-cos2θ+1)²]
=2[(sin2θ+1)²-cos²2θ]/[(sin2θ+cos2θ+1+sin2θ-cos2θ+1)(sin2θ+cos2θ+1-sin2θ+cos2θ-1)]
=2[(sin2θ+1)²-cos²2θ]/[(2sin2θ+2)·2cos2θ]
=(sin²2θ+2sin2θ+1-cos²2θ)/[2(sin2θ+1)cos2θ]
=(2sin²2θ+2sin2θ)/[2(sin2θ+1)cos2θ]
=sin2θ(sin2θ+1)/[(sin2θ+1)cos2θ]
=sin2θ/cos2θ
=tan2θ
=2[(sin2θ+1)²-cos²2θ]/[(sin2θ+cos2θ+1+sin2θ-cos2θ+1)(sin2θ+cos2θ+1-sin2θ+cos2θ-1)]
=2[(sin2θ+1)²-cos²2θ]/[(2sin2θ+2)·2cos2θ]
=(sin²2θ+2sin2θ+1-cos²2θ)/[2(sin2θ+1)cos2θ]
=(2sin²2θ+2sin2θ)/[2(sin2θ+1)cos2θ]
=sin2θ(sin2θ+1)/[(sin2θ+1)cos2θ]
=sin2θ/cos2θ
=tan2θ
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