Выражение (sin^2t-tg^2t)*(ctg^2t/sin^2t) - вопрос №87048252222 от jf7yyf 10.05.2022 19:08

Question

Алгебра: Выражение (sin^2t-tg^2t)*(ctg^2t/sin^2t)

jf7yyf · Answer

(sin²t - tg²t) · (ctg²t/sin²t) == sin²t - sin²t/cos²t) · cos²t / (sin²t ·sin²t) == ((sin²t · cos²t - sin²t)/cos²t) · (cos²t / (sin²t ·sin²t)) = = - sin²t (1 - cos²t)/(sin²t ·sin²t) = -sin²t/sin²t = -1...