$I f \cot θ + \tan θ = 3, a n d 1 - \cos^{2} θ - α \cos θ = 0, t h e n$

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$6 α^{2} (α^{2} - 9) = 1$

$9 α^{2} (α^{2} - 6) = 1$

$9 α^{2} (6 - α^{2}) = 1$

$6 α^{2} (9 - α^{2}) = 1$

answer is C.

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Detailed Solution

$\begin{array}{l} \cot θ + \tan θ = 3 \Rightarrow \frac{\cos^{2} θ + \sin^{2} θ}{\sin θ \cos θ} = 3 \Rightarrow 3 \sin θ \cos θ = 1. \\ 1 - c o s^{2} θ - α \cos θ = 0 \Rightarrow α \cos θ = \sin^{2} θ \Rightarrow \frac{α}{3 \sin θ} = \sin^{2} θ \Rightarrow \frac{α}{3} = \sin^{3} θ . α \cos θ = \frac{1}{9 \cos^{2} θ} \Rightarrow \cos^{3} θ = \frac{1}{9 α} . \\ \sin^{2} θ + \cos^{2} θ = 1 \Rightarrow {(\frac{α}{3})}^{2 / 3} + {(\frac{1}{9 α})}^{2 / 3} = 1 \Rightarrow {(\frac{1}{9 α})}^{2 / 3} = 1 - {(\frac{α}{3})}^{2 / 3} \\ \Rightarrow {(\frac{1}{9 α})}^{2} = {[1 - {(\frac{α}{3})}^{2 / 3}]}^{3} = 1 - {(\frac{α}{3})}^{2} - 3 {(\frac{α}{3})}^{2 / 3} {(\frac{1}{9 α})}^{2 / 3} = 1 - \frac{α^{2}}{9} - \frac{1}{3} = \frac{2}{3} - \frac{α^{2}}{9} \Rightarrow \frac{1}{81 α^{2}} = \frac{6 - α^{2}}{9} \\ \Rightarrow 9 α^{2} (6 - α^{2}) = 1. \end{array}$