Let -π6 β1 and α2>β2, then α1+β2 equals

x2−2xsecθ+1=0x−secθ2=tan2θx−secθ=±tanθx=secθ±tanθ α1=secθ-tanθ since α1>β1β1=secθ+tanθx2+2xtanθ+1=0x+tanθ2=sec2θx+tanθ=±secθx=−tanθ±secθ α2=−tanθ+secθ since α2>β2β2=−tanθ−secθ α1+β2=secθ-tanθ-tanθ-secθ=-2tanθ

Let -π6 β1 and α2>β2, then α1+β2 equals

x2−2xsecθ+1=0x−secθ2=tan2θx−secθ=±tanθx=secθ±tanθ α1=secθ-tanθ since α1>β1β1=secθ+tanθx2+2xtanθ+1=0x+tanθ2=sec2θx+tanθ=±secθx=−tanθ±secθ α2=−tanθ+secθ since α2>β2β2=−tanθ−secθ α1+β2=secθ-tanθ-tanθ-secθ=-2tanθ

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

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$2 (\sec θ - \tan θ)$

$2 \sec θ$

$0$

$- 2 \tan θ$

answer is D.

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Detailed Solution

$\begin{matrix} x^{2} - 2 x \sec θ + 1 = 0 \\ {(x - \sec θ)}^{2} = \tan^{2} θ \\ x - \sec θ = \pm \tan θ \\ x = \sec θ \pm \tan θ \\ α_{1} = \sec θ - \tan θ (\sin c e α_{1} > β_{1}) \\ β_{1} = \sec θ + \tan θ \end{matrix}$

$\begin{matrix} x^{2} + 2 x \tan θ + 1 = 0 \\ {(x + \tan θ)}^{2} = \sec^{2} θ \\ x + \tan θ = \pm \sec θ \\ x = - \tan θ \pm \sec θ \\ α_{2} = - \tan θ + \sec θ (\sin c e α_{2} > β_{2}) \\ β_{2} = - \tan θ - \sec θ \end{matrix}$