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Question

Let sec2(π9)+sec2(2π9)+sec2(4π9)=S and 89k=1cos6(k)=ab, where a,b are coprime, then which of the following is/are correct?

A
Number of positive divisors of a+b+S is 4.
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B
Number of positive divisors of a+b+S is 8
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C
S is a perfect square
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D
a+b is a prime number.
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Solution

The correct option is D a+b is a prime number.
S=sec2(π9)+sec2(2π9)+sec2(4π9)

=cos2 2π9 cos2 4π9+cos2 4π9 cos2 π9+cos2 π9 cos2 2π9(cos π9 cos 2π9 cos 4π9)2

=14⎢ ⎢ ⎢(cos 2π3+cos 2π9)2+(cos 5π9+cos π3)2+(cos π3+cos π9)2(164)⎥ ⎥ ⎥

=16⎢ ⎢ ⎢34+cos2 π9+cos2 2π9+cos2 4π9+cos π9cos 2π9cos 4π9(0)⎥ ⎥ ⎥

=16(34+32)=36

89k=1cos6(k)=89k=1sin6(k)=12(89k=1(sin6(k)+cos6(k)))

=12(89k=1(134sin2 (2k)))

=8923889k=1sin2(2k)

=89238×45=2218(89k=1sin2(2k)=90k=1sin2(2k)=90k=1cos2(2k)=1290k=1(sin2(2k)+cos2(2k))=45

a=221,b=8

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