证明：tanθ与tan(π/4-θ)是方程x^2+px+q=0的两个根,则
tanθ+tan(π/4-θ)=-p,tanθtan(π/4-θ)=q
因此,tan[θ+(π/4-θ)]=[tanθ+tan(π/4-θ)]/[1-tanθtan(π/4-θ)]
即tanπ/4=-p/(1-q)
整理得1=-p/(1-q)
即q=p+1