Tg² (3π/4 + α), если sinα-cosα=

B=tq² (3π/4 + α) = (1 - cos2 (3π/4 + α)) / (1 + cos2 (3π/4 + α)) =

(1 - cos (3π/2 + 2α)) / (1 + cos (3π/2 + 2α)) = (1 + sin2α) / (1 - sin2α) =

(1 + sin2α) / (1 - sin2α).

определим (1 + sin2α) / (1 - sin2α) :

sinα - cosα = (2√3) / 3 ⇒ (sinα - cosα) ² = ((2√3) / 3) ²⇔sin²α-2sinα*cosα+cos²α=4/3

1 - sin2α = 4/3 ⇒ 1-1/3 = 1 + sin2α.

B = (1 + sin2α) / (1 - sin2α) = (2/3) / (4/3) = 0,5.

ответ: 0,5.