Выражение( tg(π+α)/1-ctg²(π/2+α))·(1-tg²(3π/2-α)/ctg - вопрос №350804512132 от Govnyuk 29.12.2021 16:13

Question

Алгебра: Выражение( tg(π+α)/1-ctg²(π/2+α))·(1-tg²(3π/2-α)/ctg α)

Govnyuk · Answer

Если я нигде не ошибся:
( tg(π+α)/1-ctg²(π/2+α))·(1-tg²(3π/2-α)/ctg α) = ( tg(α)/(1-tg²(α))*(1-ctg²(α))/ctg(α)) = tg²(α)*(1-ctg²(α))/(1-tg²(α))=tg²(α)*(tg²(α)-1)/(1-tg²(α)*tg²(α)= -1.