Question

sin^6a+cos^6a+3-3cos^2a/1+tg^2a решите,пожалуйста

Answer 1

$sin ^{6} alpha +cos ^{6} alpha + frac{3-3cos ^{2} alpha}{1+tg ^{2} alpha }= \ =(sin ^{2} alpha)^{3} +(cos ^{2} alpha)^{3} + frac{3(1-cos ^{2} alpha)}{1+ frac{sin ^{2} alpha }{cos ^{2} alpha } }=$
$=(sin^ {2} alpha +cos ^{2} alpha)(sin^{4} alpha -sin^{2} alpha cdot cos ^{2} alpha + cos ^{4} alpha ) +frac{3sin ^{2} alphacdot cos ^{2} alpha }{cos ^{2} alpha + sin ^{2} alpha }= \ = sin ^{4} alpha +cos ^{4} alpha -sin^{2} alpha cdot cos ^{2} alpha + 3sin ^{2} alpha cdot cos ^{2}alpha = \= sin ^{4} alpha +cos ^{4} alpha +2sin^{2} alpha cdot cos ^{2} alpha = \(sin ^{2} alpha +cos ^{2} alpha) ^{2} =1$