If α and β are roots of the quadratic equation, x2+xsinθ−2sinθ=0,θ∈(0,π2) then α12+β12(α−12+β−12).(α−β)24 is equal to :

α+β=−sinθαβ=−2sinθNow,α12+β12(1α12+1β12).(α−β)24=((αβ)12(α+β)2−4αβ)=(−2sinθsin2θ+8sinθ)12=212(sinθ+8)12

If α and β are roots of the quadratic equation, x2+xsinθ−2sinθ=0,θ∈(0,π2) then α12+β12(α−12+β−12).(α−β)24 is equal to :

α+β=−sinθαβ=−2sinθNow,α12+β12(1α12+1β12).(α−β)24=((αβ)12(α+β)2−4αβ)=(−2sinθsin2θ+8sinθ)12=212(sinθ+8)12

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If $α and β$ are roots of the quadratic equation, $x^{2} + x \sin θ - 2 \sin θ = 0, θ \in (0, \frac{π}{2})$ then $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) . {(α - β)}^{24}}$ is equal to :

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$\frac{2^{12}}{{(\sin θ - 8)}^{6}}$

$\frac{2^{6}}{{(\sin θ + 8)}^{12}}$

$\frac{2^{12}}{{(\sin θ + 8)}^{12}}$

$\frac{2^{12}}{{(\sin θ - 4)}^{12}}$

answer is C.

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Detailed Solution

$α + β = - \sin θ$

$α β = - 2 \sin θ$

$Now, \frac{α^{12} + β^{12}}{(\frac{1}{α^{12}} + \frac{1}{β^{12}}) . {(α - β)}^{24}} = (\frac{{(α β)}^{12}}{{(α + β)}^{2} - 4 α β}) = {(\frac{- 2 \sin θ}{\sin^{2} θ + 8 \sin θ})}^{12} = \frac{2^{12}}{{(\sin θ + 8)}^{12}}$

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If α and β are roots of the quadratic equation, x2+xsinθ−2sinθ=0,θ∈(0,π2) then α12+β12(α−12+β−12).(α−β)24 is equal to :

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If $α and β$ are roots of the quadratic equation, $x^{2} + x \sin θ - 2 \sin θ = 0, θ \in (0, \frac{π}{2})$ then $\frac{α^{12} + β^{12}}{(α^{- 12} + β^{- 12}) . {(α - β)}^{24}}$ is equal to :