Given Tan (a + π / 4) = - 1 / 2, π / 2 & lt; a & lt; π, find the value of (sin2a-2cos ^ 2a) / √ 2Sin (a - π / 4)

Given Tan (a + π / 4) = - 1 / 2, π / 2 & lt; a & lt; π, find the value of (sin2a-2cos ^ 2a) / √ 2Sin (a - π / 4)

Tan2a = - 2 √ 2; tan2a = 2tana / (1-tana * Tana) = - 2 √ 2. By simplifying Tana * Tana - (√ 2 / 2) * Tana - 1 = 0, Tana = - 2 / 2 or Tana = √ 2. Because 2A belongs to (π / 2, π), a belongs to (π / 4, π / 2), so Tana = √ 2 (2cos square A / 2-sina-1) / (√ 2S