1) 1-9x²=0 2) 16+3x²=8x 3) 18-x²+3x=0 - id1082728920201229 от вау39 29.12.2020 11:39

Question

Алгебра: 1) 1-9x²=0 2) 16+3x²=8x 3) 18-x²+3x=0

вау39 · Accepted Answer

1)
1/cos^2(a) – tg^2(a)-sin^2(a)= 1/cos^2(a) – sin^2(a)/cos^2(a))-sin^2(a)=
=[1 – sin^2(a) ]/cos^2(a))-sin^2(a)=cos^2(a)/cos^2(a))-sin^2(a)=
=1-sin^2(a)=cos^2(a)
2)
cos^2(a)+ctg^2(a)-1/sin^2(a)=cos^2(a)+[cos^2(a)-1]/sin^2(a)=
=cos^2(a)-sin^2(a)]/sin^2(a)=cos^2(a)-1 = -sin^2(a)
3)
1/cos^2(a) – tg^2(a)(cos^2(a)+1)=1/cos^2(a) – sin^2(a)-sin^2(a)/(cos^2(a)=
=(1 – sin^2(a))/cos^2(a)-sin^2(a)=1-sin^2(a)=cos^2(a)

4) (1+sin^2(a))ctg^2(a) – 1/sin^2(a)=cos^2/sin^2 +cos^2 – 1/sin^2(a)=
=(cos^2 - 1)/sin^2 +cos^2= -sin^2/sin^2 +cos^2= -1+cos^2= -sin^2(a)

5)sin(a)/(1+cos(a)) + sin(a)/(1-cos(a))=sin(a) * ((1-cosa)+(1+cosa)) / (1-cos^2)=2/sin(a)

6)cos(a)/(1+sin(a))+ cos(a)/(1-sin(a))=cos(a)* ((1-sina+1+sina)) / (1-sin^2a)= 2/cos(a)

1) 1-9x²=02) 16+3x²=8x3) 18-x²+3x=0

MOGZ ответил

1) 1-9x²=0

2) 16+3x²=8x

3) 18-x²+3x=0