Question

Ребятки прошу вас помоочь ,с 5 задания *.*

Answer 1

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Answer 2

6) sin^4β+sin²β*cos²β=sin²β(sin²β+cos²β)=sin²β
7) tg²β*ctg²β-sin²β=1-sin²β=cos²β
8) 1-cos²β/sin²β-1=1-cos²β/-(1-sin²β)=-sin²β/cos²β=-tg²β
9) cosα+ctgα/1+sinα=cosαsinα+cosα/sinα*1/1+sinα=cosα(1+sinα)/sinα*1/1+sinα=cosα/sinα=ctgα
10) (1-cos²α)(1+tg²α)=sin²α*1/cos²α=sin²α/cos²α=tg²α
11) (sinα/1+cosα)+(1+cosα/sinα)=(sin²α+(1+cosα)²)/(sinα(1+cosα))=(sin²α+cos²α+2cosα+1)/(sinα(1+cosα))=2+2cosα/sinα(1+cosα)=2(1+cosα)/sinα(1+cosα)=2/sinα
12) (cosα/1+sinα)-(cosα/1-sinα)=(cosα(1-sinα)-cosα(1+sinα))/1-sin²α=-2cosαsinα/cos²α=-2sinα/cosα=-2tgα
13) 1-2sinαcosα/sinα-cosα=sin²α-2sinαcosα+cos²α/sinα-cosα=(sinα-cosα)²/sinα-cosα=sinα-cosα
14) sin^4α-cos^4α+cos²α=(sin²α+cos²α)(sin²α-cos²α)+cos²α=sin²α-cos²α+cos²α=sin²α
15) 1+ctg^4α/ctg²α+tg²α=(sin^4α+cos^4α/sin^4α)*(sin²αcos²α/cos^4α+sin^4α)=cos²α/sin²α=ctg²α
16) (cos²α/1+tg²α)-(sin²α/1+ctg²α)=(cos²α*cos²α)-(sin²α*(-sin²α))=cos^4α+sin^4α

Answer 3

Спасибо:)