4Sinα-3cosα __________ 5sinα Если: а) tgα=2; b) ctgα=3
А) tg(a) = sin(a)/cos(a); tg(a) = 2; sin(a) = 2cos(a); cos(a) = x; sin(a) = 2x; (3*2x - 3x)/(5*2x) = 3x/10x = 3/10. b) ctg(a) = cos(a)/sin(a); ctg(a) = 3; cos(a) = 3sin(a); sin(a) = y; cos(a) = 3y; (3y - 3*3y)/5y = -6y/5y = -6/5.