(cos^2 t - ctg^2)/(sin^2 t -th^2 t) - вопрос №22224208065 от 287888 10.04.2020 03:01

Question

Математика: (cos^2 t - ctg^2)/(sin^2 t -th^2 t)

287888 · Answer

(cos²t-ctg²t)/(sin²t-tg²t)=ctg⁶t1. cos²t-ctg²t=cos²t-cos²t/sin²t=(cos²t *sin²t-cos²t)/sin²t=cos²t*(sin²t-1)/sin²t=-cos⁴t/sin²t2. sin²t-tg²t=sin²t-sin²t/cos²t=(sin²t *cos²t-sin²t)/cos²t=sin²t*(cos²t-1)/cos²t=-sin⁴t/cos²t3. (-cos⁴t/sin²t)/(-sin⁴t/cos²t)=cos⁶t/sin⁶t=ctg⁶t...