Question

Докажите тождество 1) 1+2sinacosa / (sina+cosa)^2 =1 2) sin^2a-cos^2a+1 / sin^2a=2 3) (2-sina)(2+sina)+(2-cosa)(2+cosa)=7

Answer 1

1) 1+2sinacosa / (sina+cosa)^2 =(sin²a+cos²a+2sinacosa)/(sin²a+2sinacosa+cos²a)=1
2) sin^2a-cos^2a+1 / sin^2a=(sin²a-cos²a+sin²a+cos²a)/sin²a=2sin²a/sin²a=2
3) (2-sina)(2+sina)+(2-cosa)(2+cosa)=4-sin²a+4-cos²a=8-(sin²a+cos²a)=8-1=7

Answer 2

$3) (2-sin \alpha)(2+sin \alpha)+ (2-cos \alpha)(2+cos \alpha)=7 \\ 4-sin^{2} \alpha +4-cos^{2} \alpha =7 \\ 8-(sin^{2} \alpha +cos^{2} \alpha )=7 \\ 8-1=7 \\ 7=7$$1) \frac{1+2sin \alpha cos \alpha }{(sin \alpha +cos \alpha )^{2}}=1 \\ \frac{1+2sin \alpha cos \alpha }{(sin^{2} \alpha +2sin \alpha cos \alpha + cos^{2} \alpha )}=1 \\ \frac{1+2sin \alpha cos \alpha }{(1 +2sin \alpha cos \alpha)}=1 \\ 1=1 \\ 2) \frac{sin^{2} \alpha -cos^{2} \alpha +1}{sin^{2} \alpha } =2 \\ \frac{sin^{2} \alpha -cos^{2} \alpha +sin^{2} \alpha +cos^{2} \alpha}{sin^{2} \alpha } =2 \\ \frac{2sin^{2} \alpha }{sin^{2} \alpha } =2 \\ 2=2$