Arctg √3 ctg a, если cos a= -4/5 и π/2<а<π

Question

Arctg √3
ctg a, если cos a= -4/5 и π/2<а<π

Answer 1

1) arctg√3 = π/3
2) Sinα = √(1 - Cos²α = √( 1 -  16/25) = √9/25 = 3/5
ctgα = Cosα/Sinα = - 4/5 : 3/5 = - 4/5 * 5/3 = - 4/3 = - 1 1/3