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

x2−2xsecθ+1=0⇒x=secθ±tanθ and−π6 secθ>sec(−π12)⇒sec(π6)>secθ>sec(π12)andtan(−π6) −tanθ>tan(π12)α,βare the root of x2−2xsecθ+1=0 and α1>β1⇒α1=secθ−tanθ and β1=secθ+tanθ⇒α2β2are the root of x2+2xtanθ−1=0 and α2>β2α2=−tanθ+secθ,β2=−tanθ−secθHere α1+β2=−2tanθ

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

x2−2xsecθ+1=0⇒x=secθ±tanθ and−π6 secθ>sec(−π12)⇒sec(π6)>secθ>sec(π12)andtan(−π6) −tanθ>tan(π12)α,βare the root of x2−2xsecθ+1=0 and α1>β1⇒α1=secθ−tanθ and β1=secθ+tanθ⇒α2β2are the root of x2+2xtanθ−1=0 and α2>β2α2=−tanθ+secθ,β2=−tanθ−secθHere α1+β2=−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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$2 \sec θ$

$- 2 \sec θ$

$- 2 \tan θ$

$2 \tan θ$

answer is C.

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Detailed Solution

$\begin{array}{l} x^{2} - 2 x \sec θ + 1 = 0 \\ \Rightarrow x = \sec θ \pm \tan θ a n d - \frac{π}{6} < θ < - \frac{π}{12} \\ \Rightarrow \sec (- \frac{π}{6}) > \sec θ > \sec (- \frac{π}{12}) \\ \Rightarrow \sec (\frac{π}{6}) > \sec θ > \sec (\frac{π}{12}) \\ a n d \tan (- \frac{π}{6}) < \tan θ < \tan (- \frac{π}{12}) \\ \Rightarrow \tan (\frac{π}{6}) > - \tan θ > \tan (\frac{π}{12}) \\ α, β a r e t h e r o o t o f x^{2} - 2 x \sec θ + 1 = 0 a n d α_{1} > β_{1} \\ \Rightarrow α_{1} = \sec θ - \tan θ a n d β_{1} = \sec θ + \tan θ \\ \Rightarrow α_{2} β_{2} a r e t h e r o o t o f x^{2} + 2 x \tan θ - 1 = 0 a n d α_{2} > β_{2} \\ α_{2} = - \tan θ + \sec θ, β_{2} = - \tan θ - \sec θ \\ H e r e α_{1} + β_{2} = - 2 \tan θ \end{array}$