Question

If ,
$cos \alpha + sec \alpha = 2 \\ \\ find \: the \: value \: of \: \\ {cos \alpha }^{2018} + {sec \alpha }^{2018}$

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Answer 1

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Here is your answer

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$cos \alpha + sec \alpha = 2 \\ \\ cos \alpha + \frac{1}{cos \alpha } = 2 \\ \\ \frac{ {cos \alpha }^{2} + 1}{cos \alpha } = 2 \\ \\ {cos \alpha }^{2} + 1 = 2cos \alpha \\ \\ {cos \alpha }^{2} - 2cos \alpha + 1 = 0 \\ cos \alpha = 1 \\ \\ then \: \: \frac{1}{cos \alpha } = sec \alpha = 1 \\ \\ then \\ {cos \alpha }^{2018} + {sec \alpha }^{2018} \\ \\ = 1 + 1 \\ = 2$


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Answer 2

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