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Question:

Let−π6<θ<−π12, Supposeα1andβ1are the roots of the equationx2𕒶xsecθ+1=0andα2andβ2are the roots of the equationx2+2xtanθ𕒵=0. Ifα1>β1andα2>β2, thenα1+β2equals to2secθ𕒶tanθ02(secθ−tanθ)Let−π6<θ<−π12, Supposeα1andβ1are the roots of the equationx2𕒶xsecθ+1=0andα2andβ2are the roots of the equationx2+2xtanθ𕒵=0. Ifα1>β1andα2>β2, thenα1+β2equals to2secθ𕒶tanθ02(secθ−tanθ)−π6<θ<−π12−π6<θ<−π12−π6<θ<−π12−π6<θ<−π12−π6<θ<−π12−−π6π6πππ666<<θθ<<−−π12π12πππ121212α1α1α1α1α1α1ααααα11111β1β1β1β1β1β1βββββ11111x2𕒶xsecθ+1=0x2𕒶xsecθ+1=0x2𕒶xsecθ+1=0x2𕒶xsecθ+1=0x2𕒶xsecθ+1=0x2xxxxx22222−𕒶2xxsecsecθθθθ++11==00α2α2α2α2α2α2ααααα22222β2β2β2β2β2β2βββββ22222x2+2xtanθ𕒵=0x2+2xtanθ𕒵=0x2+2xtanθ𕒵=0x2+2xtanθ𕒵=0x2+2xtanθ𕒵=0x2xxxxx22222++22xxtantanθθθθ−𕒵1==00α1>β1α1>β1α1>β1α1>β1α1>β1α1ααααα11111>>β1βββββ11111α2>β2α2>β2α2>β2α2>β2α2>β2α2ααααα22222>>β2βββββ22222α1+β2α1+β2α1+β2α1+β2α1+β2α1ααααα11111++β2βββββ222222secθ2secθ2secθ2secθ2secθ2secθ22secsecθθθθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ−𕒶2tantanθθθθ0000002(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)22(secθ−tanθ)((secsecθθθθ−−tantanθθθθ))A2secθ2secθ2secθ2secθ2secθ2secθ2secθ22secsecθθθθB0000000C2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)2(secθ−tanθ)22(secθ−tanθ)((secsecθθθθ−−tantanθθθθ))D𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ𕒶tanθ−𕒶2tantanθθθθ?

2secθ

0

2(secθ−tanθ)

𕒶tanθ

Solution:

x2𕒶xsecθ+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