Question

Помогите доказать tg2atga/tg2a-tga=sin2a

Answer 1

(sin2a/cos2a)(sina/cosa)/((sin2a/cos2a)-(sina/cosa))==(sin2asina/cos2acosa)/(sin2acosa-sinacos2a/cos2acosa)==sin2asina/(sin2acosa-sinacos2a)=sina/(cosa-(sinacos2a/sin2a))==sina/(cosa-(cos2a/2cosa))=2sinacosa/2cos.^2a-cos2a==2sinacosa/2cos.^2-cos.^2+sin.^2=2sinacosa/cos.^2a+sin.^2a=2sinacosa=sin2a