cyklometryczna Gris: oblicz arctg(x+2)−arctg(x+1)=π/4 Jak sie za takie cos zabiera? dziekuje

Trivial:

	tgα + tgβ
tg(α+β) =
	1 − tgα*tgβ

	π
arctg(x+2) − arctg(x+1) =		/ tg
	4

α = arctg(x+2), tgα = x+2 β = −arctg(x+1), tgβ = −(x+1)

x+2 + [−(x+1)]		π
	= tg(		)
1 − (x+2)*[−(x+1)]		4

1
	= 1 / * (1 + x2+3x+2)
1 + x2+3x+2

1 = 1 + x²+3x+2 x² + 3x + 2 = 0 Δ = 9 − 8 = 1; √Δ = 1. x₁ = −2, x₂ = −1.