Тригонометрия, упростить выражения! Помогите пожалуйста.

a)2sin²a б)(1+сosa-cos²a)/cosa=(sin²a+cosa)/cosa в)cos2a+cos2a=2cos2a д)2sin²acosa/sina=2sinacosa=sin2a е)sin²a-2sinacosa+cos²a=1-sin2a 2 sin²a+cos²a+2sinacosa)/(sina+cosa)=(sina+cosa)²/(sina+cosa)=sina+cosa

№1. а) 1-cos(2*α)=sin²(α)+cos²(α)-(cos²(α)-sin²(α))=2*sin²(α) б) (1+cos(α))/(cos(α))-cos(α)=(1+cos(α)-cos²(α))/(cos(α))=(sin²(α)+cos(α))/cos(α)=1+sin²(α)/cos(α) в) 1-2*sin²(α)+cos(2*α)=sin²(α)+cos²(α)-2*sin²(α)+cos²(α)-sin²(α)=2*cos²(α)-2*sin²(α)=2*(cos²(α)-sin²(α))=2*cos(2*α) д) (1-cos(2*α))*ctg(α)=(sin²(α)+cos²(α)-(cos²(α)-sin²(α))*cos(α)/sin(α)=2*sin²(α)*cos(α)/sin(α)=2*sin(α)*cos(α)=sin(2*α) е) (sin(α)-cos(α))²=sin²(α)-2*sin(α)*cos(α)+cos²(α)=1-2*sin(α)*cos(α)=1-sin(2*α) №2. (1+sin(2*α))/*(sin(α)+cos(α))=(sin²(α)+2*sin(α)*cos(α)+cos²(α))/(sin(α)+cos(α))=(sin(α)+cos(α))²/((sin(α)+cos(α))=sin(α)+cos(α). Тождество доказано.