Решение
1) ctg²α + cos²α - 1/sin²α = cos²α/sin²α - 1/sin²α + cos²α =
= - (1 - cos²α)/sin²α + cos²α = - sin²α/sin²α + cos²α =
= - 1 + cos²α = - (1 - cos²α) = - sin²α
или
ctg²α + cos²α - 1/sin²α = ctg²α + cos²α - (сtg²α + 1) =
= ctg²α + cos²α - сtg²α - 1 = - (1 - cos²α) = - sin²α
2) cos(π/2 + x)*cosx + sin(π/2 + x)*sinx = 0
Упростим левую часть
cos(π/2 + x)*cosx + sin(π/2 + x)*sinx = - sinx*cosx + cosx*sinx = 0
0 = 0
доказано