Question

Помогите кто какой решит

Answer 1

Ответ:1+tg²α=1/cos²α  и cos²α(1/cos²α)-1=1-1=0

1- sin²α(1+ctg²α)=1-sin²α·1/sin²α=1-1=0

1+tg²α+1/sin²α=1+tg²α+1+tg²α=2(1+tg²α)

(1+tg²α)/(1+ctg²α)=1/cos²α:1/sin²α=sin²α/cos²α=ctg²α

(sin²α-cos²α)²+sin²αcos²α=sin⁴α-2sin²αcos²α+cos⁴α

+2sin²αcos²α=sin⁴α+cos⁴α

1/(tgα+ctgα)=1/(sinα/cosα+cosα/sinα)=1/(sin²α+cos²α)/sinα·cosα= sinα·cosα/(sin²α+cos²α)=sinα·cosα=0.5·2·sinα·cosα=0.5sin2α