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Let $\frac{- π}{6} < θ < \frac{- π}{12}$ . Suppose $α_{1}, β_{1}$ are roots of $x^{2} - 2 x \sec θ + 1 = 0$ and $α_{2}, β_{2}$ are roots of $x^{2} + 2 x \tan θ - 1 = 0$ If $α_{1} < β_{1} a n d α_{2} < β_{2}$ then $α_{1} + β_{2} =$

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$2 (\sec θ - \tan θ)$

$2 \sec θ$

$- 2 \tan θ$

answer is B.

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Detailed Solution

$α_{1} β_{1} = \sec θ \pm T a n θ a n d α_{1} < β_{1} \Rightarrow α_{1} = S e c θ + T a n θ α_{2}, β_{2} = - \tan θ \pm \sec θ a n d α_{2} < β_{2} \Rightarrow β_{2} = - T a n θ + \sec θ α_{1} + β_{2} = 2 \sec θ$

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