Question

sin2αtanα+cos2α/tanα+2sinαcosα化简

这是对的~
cos2α/tanα是一个整体

Answer 1

(sin2αtanα+cos2α)/(tanα+2sinαcosα)
分数线上下都乘以cosα得
=(sin2α·sinα+cos2α·cosα)/(sinα+2sinα·cos²α)
=(cos2α·cosα+sin2α·sinα)/(sinα+2sinα·cos²α)
=sin(2α-α) / (sinα+2sinα·cos²α)
=sinα / (sinα+2sinα·cos²α)
=1 / (1+2cos²α)

Answer 2

化简 (sin2αtanα+cos2α)/(tanα+2sinαcosα)
=(sin2a*sina/cosa+cos2a)/(sina/cosa+2sinacosa)
=(sin2a*sina+cos2a*cosa)/(sina+2sina*cos^2a)
=cos(2a-a)/sina(a+2cos^2a)
=cosa/sina(a+2cos^2a)

Answer 3

（sin2αtanα+cos2α）/（tanα+2sinαcosα）
=（sinαsin2α+cosαcos2α)/(sinα+2sinαcos^α)
=cosα/[sinα(2+cos2α)],

sin2αtanα+cos2α/tanα+2sinαcosα
=sinαsin2α/cosα+cosαcos2α/sinα+sin2α
=2sin^α+(cosαcos2α+sinαsin2α)/sinα
=2sin^α+cosα/sinα
=1-cos2α+cotα.

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