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Q.

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

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a

2(sec⁡θ−tan⁡θ)

b

2sec⁡θ

c

−2tan⁡θ

d

0

answer is C.

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Detailed Solution

Solving given equations and from given values of θ which lie in fourth quadrantα1=sec⁡θ−tan⁡θβ1=sec⁡θ+tan⁡θα2=sec⁡θ−tan⁡θβ2=−sec⁡θ−tan⁡θα1+β2=−2tan⁡θ
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Let −π6<θ<−π12. Suppose α1 and β1 are the roots of the equation x2−2xsec⁡θ+1=0 and α2 and β2 are the roots of the equation x2+2xtan⁡θ−1=0.  If α1>β1 and α2>β2, then α1+β2 equals