Let −π6 β1 and α2>β2 then α1+β2 equals

x2−2xsec⁡θ+1=0⇒x=sec⁡θ±tan⁡θ and −π6 sec⁡θ>sec⁡−π12⇒sec⁡π6>sec⁡θ>sec⁡π12 and tan⁡−π6 −tan⁡θ>tan⁡π12∵α,β, are the root of x2−2xsec⁡θ+1=0 and α1>β1 ⇒α1=sec⁡θ−tan⁡θ and β1=sec⁡θ+tan⁡θ⇒α2β2 are the roots of x2+2xtan⁡θ−1=0 and α2>β2∴α2=−tan⁡θ+sec⁡θ,β2=−tan⁡θ−sec⁡θHere α1+β2=−2tan⁡θ.

Let −π6 β1 and α2>β2 then α1+β2 equals

x2−2xsec⁡θ+1=0⇒x=sec⁡θ±tan⁡θ and −π6 sec⁡θ>sec⁡−π12⇒sec⁡π6>sec⁡θ>sec⁡π12 and tan⁡−π6 −tan⁡θ>tan⁡π12∵α,β, are the root of x2−2xsec⁡θ+1=0 and α1>β1 ⇒α1=sec⁡θ−tan⁡θ and β1=sec⁡θ+tan⁡θ⇒α2β2 are the roots of x2+2xtan⁡θ−1=0 and α2>β2∴α2=−tan⁡θ+sec⁡θ,β2=−tan⁡θ−sec⁡θHere α1+β2=−2tan⁡θ.

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Let $- \frac{π}{6} < θ < \frac{- π}{12}$ . Suppose α₁ and ß₁ are the roots of the equation $x^{2} - 2 xsec θ + 1 = 0$ and α₂ and ß₂ are the roots of the equation $x^{2} + 2 xtan θ - 1 = 0$ . If $α_{1} > β_{1} and α_{2} > β_{2} then α_{1} + β_{2}$ equals

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$- 2 \sec θ$

$2 \sec θ$

$- 2 \tan θ$

$2 \tan θ$

answer is C.

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Detailed Solution

$\begin{array}{l} x^{2} - 2 xsec θ + 1 = 0 \\ \Rightarrow x = \sec θ \pm \tan θ and - \frac{π}{6} < θ < - \frac{π}{12} \\ \Rightarrow \sec (- \frac{π}{6}) > \sec θ > \sec (- \frac{π}{12}) \\ \Rightarrow \sec (\frac{π}{6}) > \sec θ > \sec (\frac{π}{12}) \\ and \tan (- \frac{π}{6}) < \tan θ < \tan (- \frac{π}{12}) \\ \Rightarrow \tan (\frac{π}{6}) > - \tan θ > \tan (\frac{π}{12}) \end{array}$
$∵ α, β, are the root of x^{2} - 2 xsec θ + 1 = 0 and α_{1} > β_{1} \begin{array}{l} \Rightarrow α_{1} = \sec θ - \tan θ and β_{1} = \sec θ + \tan θ \\ \Rightarrow α_{2} β_{2} are the roots of x^{2} + 2 xtan θ - 1 = 0 and α_{2} > β_{2} \\ ∴ α_{2} = - \tan θ + \sec θ, β_{2} = - \tan θ - \sec θ \end{array}$
$Here α_{1} + β_{2} = - 2 \tan θ .$