tan2θ=2√2,θ∈（π/2,π）求 [2cos²(θ/2)-sinθ-1]/(sinθ+cosθ )tan2θ=2tanθ/(1-tan²θ)=2√2tanθ=(√2)(1-tan²θ) (√2)tan²θ+tanθ-√2=0tanθ=(-1±3)/(2√2) θ∈（π/2,π）∴ta...