Tg²(3π/4 +α), если sinα-cosα=[latex] \frac{2 \sqrt{3} }3} [/latex]

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 .