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Question

If x+1x=2cosα then xn+1xn is equal to:

A
2ncosα
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B
2ncosnα
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C
2isinna
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D
2cosnα
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Solution

The correct option is A 2cosnα
Put x=cosα+isinα=eiα
So 1x=x1=eiα=cosαisinα
Clearly x+1x=2cosα
Hence xn+1xn=xn+xn=(cosα+isinα)n+(cosα+isinα)n
=cosnα+isinnα+sinnαisinnα
=2cosnα, Use Demoivre's theorem

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