Let -π6<θ<-π12. Suppose α1 and β1 are the roots of the equation x2-2xsecθ+1=0 and α2 and β2 are the roots of the equation x2+2xtanθ-1=0. If α1>β1 and α2>β2, then α1+β2 equals

x2−2xsecθ+1=0x−secθ2=tan2θx−secθ=±tanθx=secθ±tanθ                    α1=secθ-tanθ  since α1>β1β1=secθ+tanθx2+2xtanθ+1=0x+tanθ2=sec2θx+tanθ=±secθx=−tanθ±secθ                    α2=−tanθ+secθ  since α2>β2β2=−tanθ−secθ                 α1+β2=secθ-tanθ-tanθ-secθ=-2tanθ