Oblicz wartość wyrażenia: √5 (ctgα-tgα),jeśli sinα= 2/3 Sprawdź tożsamość: 1+ctgα=(sinα+cosα)/sinα

Zad. 1) Dane: sinα= 2/3 √5 (ctgα-tgα) = ? tgα = sinα/cosα ctgα = 1/tgα sin²α + cos²α = 1 cos²α = 1 - sin²α cos²α = 1 - (2/3)² cos²α = 1 - 4/9 cos²α = 5/9 cos²α - 5/9 = 0 (cosα - √5/3)(cosα + √5/3) = 0 cosα - √5/3 = 0 ∨ cosα + √5/3 = 0 cosα = √5/3 ∨ cosα = - √5/3 tgα = sinα/cosα tgα = (2/3): (√5/3) ∨ tgα = (2/3): (- √5/3) tgα = 0,4√5 ∨ tgα = - 0,4√5 ctgα = 1/0,4√5 ∨ ctgα = - 1/0,4√5 ctgα = 0,5√5 ∨ ctgα = - 0,5√5 √5 (ctgα-tgα) = √5 (0,5√5 - 0,4√5) = √5*0,1√5 = 0,5 lub 5 (ctgα-tgα) = √5 (- 0,5√5 + 0,4√5) = - √5*0,1√5 = - 0,5 Zad. 2) 1 + ctgα = (sinα + cosα)/sinα P = (sinα + cosα)/sinα = sinα/sinα + cosα/sinα = = 1 + ctgα = = P