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ГДЗ Геометрія 9 клас Бурда НУШ
Розділ 2. Координати і вектори на площині
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3) ((sin α+cosα)^2- 1)/(〖(1+tg α)〗^2- (〖1-tg α)〗^2 ) = (sin^2 α+2 sin〖α cos〖α+ cos^2 α-1〗 〗)/(1+2 tg α+ tg^2 α-(1-2tg α+ tg^2 α)) = (1+2 sin〖α cos〖α-1〗 〗)/(1+2 tg α+ tg^2 α-1+2 tg α- tg^2 α) = (2 sin〖α cosα 〗)/(4 • sinα/cosα ) = sin〖α cos^2 α〗/(2 sinα ) = 1/2 cos2 α;
4) (〖1-sin〗^4 α- cos^2 α)/(cos^2 α sin^2 α) = ((1- sin^4 α)- cos^4 α)/(cos^2 α sin^2 α) = (〖(1-sin〗^2 α)(1- sin^2 α)- cos^4 α)/(cos^2 α sin^2 α) = (cos^2 α(1+ sin^2 α)- cos^4 α)/(cos^2 α sin^2 α) = (cos^2 α(1+ sin^2 α- cos^2 α))/(cos^2 α sin^2 α) = (sin^2 α+ sin^2 α)/(sin^2 α) = (〖2sin〗^2 α)/(sin^2 α) = 2.
4) (〖1-sin〗^4 α- cos^2 α)/(cos^2 α sin^2 α) = ((1- sin^4 α)- cos^4 α)/(cos^2 α sin^2 α) = (〖(1-sin〗^2 α)(1- sin^2 α)- cos^4 α)/(cos^2 α sin^2 α) = (cos^2 α(1+ sin^2 α)- cos^4 α)/(cos^2 α sin^2 α) = (cos^2 α(1+ sin^2 α- cos^2 α))/(cos^2 α sin^2 α) = (sin^2 α+ sin^2 α)/(sin^2 α) = (〖2sin〗^2 α)/(sin^2 α) = 2.