Let −π$$6$$ β$$1$$ and α$$2$$>β$$2$$, then α$$1+$$β$$2$$ equals

Correct option is 3−2tan⁡θ

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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2(sec⁡θ−tan⁡θ)

2sec⁡θ

−2tan⁡θ

answer is C.

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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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