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$[1+\sec 2 \theta][1+\sec 4 \theta]=$
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The correct answer is:
$4 \cot \theta \tan 4 \theta$
$[1+\sec 2 \theta][1+\sec 4 \theta]=\left[1+\frac{1}{\cos 2 \theta}\right] \cdot\left[1+\frac{1}{\cos 4 \theta}\right]$
$\begin{aligned} & =\frac{2 \cos ^2 \theta}{\cos 2 \theta} \cdot \frac{2 \cos ^2 2 \theta}{\cos 4 \theta} \\ & =\frac{\sin 2 \theta \cdot \cos \theta \cdot 2 \cos 2 \theta}{\sin \theta \cdot \cos 4 \theta} \\ & =\frac{\cos \theta \cdot \sin 4 \theta}{\sin \theta \cos 4 \theta}=\cot \theta \cdot \tan 4 \theta\end{aligned}$
$\begin{aligned} & =\frac{2 \cos ^2 \theta}{\cos 2 \theta} \cdot \frac{2 \cos ^2 2 \theta}{\cos 4 \theta} \\ & =\frac{\sin 2 \theta \cdot \cos \theta \cdot 2 \cos 2 \theta}{\sin \theta \cdot \cos 4 \theta} \\ & =\frac{\cos \theta \cdot \sin 4 \theta}{\sin \theta \cos 4 \theta}=\cot \theta \cdot \tan 4 \theta\end{aligned}$
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