Question

一道三角函数化简题

需要过程，最好也上传图片回答，谢谢各位！
答案：1

Answer 1

看图片吧

Answer 2

(1-2sinacosa)/(cosa^2-sina^2)*(1+2sinacosa)/(1-2sina^2)
=(1-sin2a)/cos2a*(1+sin2a)/cos2a
=(1-sin2a^2)/cos2a^2
=cos2a^2/cos2a^2
=1

Answer 3

(sin²a+cos²a-2sinacosa)/(cosa+sina)(cosa-sina)*【(sin²a+cos²a+2sinacosa)/(sin²a+cos²a-2sin²a)】
=(sina-cosa)²/(cosa+sina)(cosa-sina)*【(sina+cosa)²/(cosa-sina)（cosa+sina）】
=(cosa-sina)/(cosa+sina)*[(sina+cosa)/(cosa-sina)]
=1

Answer 4

(1-2sinαcosα)/[(cosα)^2-(sinα)^2]*(1+2sinαcosα)/[1-2(sinα)^2]
=(1-sin2α)/(cos2α)*(1+sin2α)/(cos2α)
=(1-sin2α)(1+sin2α)/(cos2α)^2
=[1-(sin2α)^2/(cos2α)^2
=(cos2α)^2/(cos2α)^2
=1