Question

ДОКАЗАТЬ ТОЖДЕСТВО
(sinAcosA)/ctgA = 1 - (ctg^2A - cos^2A)/ctg^2A

Answer 1

(sinAcosA)/ctgA = 1 - (ctg²A - cos²A)/ctg²A
(sinAcosA)/(cosA/sinA) = (ctg²A - ctg²A  + cos²A)/ctg²A
sin²AcosA/cosA = cos²A/ctg²A
sin²A = cos²A/(cos²A/sin²A)
sin²A = cos²Asin²A/cos²A
sin²A = sin²A

Answer 2

помогите пожалуйста еще с этим https://znanija.com/task/25158776

Answer 3

и этим https://znanija.com/task/25158848

Answer 4

(sinAcosA)/ctgA = 1 - (ctg^2A - cos^2A)/ctg^2A
sinAcosA/ctgA=sinAcosA:cosA/sinA=sinAcosA*sinA/cosA=sin²A
1 - (ctg^2A - cos^2A)/ctg^2A=1-(cos²A/sin²A-cos²A):cos²A/sin²A=
=1-cos²A(1-sin²A)/sin²A *sin²A/cos²A=1-1+sin²A=sin²A
sin²A=sin²A