tg²(3π/4 +α), если sinα-cosα=[tex] \frac{2 \sqrt{3} }3} [/tex]

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 .