[(tana*tan2a)/(tan2a-tana)]+√3(sin ^2 a-cos^2 a)=2sin [2a-(pi/3)]要过程,

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[(tana*tan2a)/(tan2a-tana)]+√3(sin ^2 a-cos^2 a)=2sin [2a-(pi/3)]要过程,

[(tana*tan2a)/(tan2a-tana)]+√3(sin ^2 a-cos^2 a)=2sin [2a-(pi/3)]要过程,
[(tana*tan2a)/(tan2a-tana)]+√3(sin ^2 a-cos^2 a)=2sin [2a-(pi/3)]要过程,

[(tana*tan2a)/(tan2a-tana)]+√3(sin ^2 a-cos^2 a)=2sin [2a-(pi/3)]要过程,
这个题目的关键是处理(tana*tan2a)/(tan2a-tana)
把tana=sina/cosa,tan2a=sin2a/cos2a
代入得
(tana*tan2a)/(tan2a-tana)
=sinasin2a/(cosaasin2a-sinacos2a)
=sinasin2a/sin(2a-a)
=sin2a
后面的想必你会了

原式=sin2a*sina/(sin2a*cosa-cos2a*sina)-√3(cos^2 a-sin ^2 a)
=sin2a*sina/sina-√3cos2a
=sin2a-√3cos2a
=2(1/2sin2a-√3/2cos2a)
=2sin (2a-π/3)