2secθ
0
2(secθ−tanθ)
tanθ
x2xsecθ+1=0has rootsα1,β1, out of whichα1>β1So,x=2secθ±2tanθ2=secθ±tanθIn the given range ofθ,secθis positive, whiletanθis negative and so,α1=secθ−tanθSimilarly, forx2+2xtanθ=0,x=tanθ±2secθ2=−tanθ±secθSinceα2>β2,β2=−tanθ−secθ⇒α1+β2=tanθ