If $\cot^{3} α + \cot^{2} α + \cot α = 1$ , then

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$\cos 2 α \cdot \tan α = 1$

$\cos 2 α \cdot \tan α = - 1$

$\cos 2 α - \tan 2 α = - 1$

$\cos 2 α - \tan 2 α = 1$

answer is A, C.

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Detailed Solution

$\begin{array}{l} \cot^{3} α + \cot^{2} α + \cot α = 1 \\ \Rightarrow \tan^{3} α - \tan^{2} α - \tan α - 1 = 0 \\ \Rightarrow \tan α (\tan^{2} α - 1) = 1 + \tan^{2} α \\ \Rightarrow \tan α = - (\frac{1 + \tan^{2} α}{1 - \tan^{2} α}) \\ \Rightarrow \cos 2 α \cdot \tan α = - 1 \end{array}$
Now $\cos 2 α - \tan 2 α = - \frac{1}{\tan α} - \frac{2 \tan α}{1 - \tan^{2} α}$
$\begin{array}{l} = - (\frac{1 - \tan^{2} α + 2 \tan^{2} α}{\tan α (1 - \tan^{2} α)}) \\ = - \frac{1 + \tan^{2} α}{\tan α (1 - \tan^{2} α)} \\ = - \frac{1}{\tan αcos 2 α} \\ = 1 \end{array}$