The real value of θ for which the expression 1+icos⁡θ1−2icos⁡θ is a real number, where i=−1 , is

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nπ+π4

nπ±π4

2nπ±π2

nπ+−1nπ4

answer is C.

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Detailed Solution

1+icos⁡θ1−2icos⁡θ=1−2cos2⁡θ+3icos⁡θ1+4cos2⁡θReal⇒cos⁡θ=0⇒θ=2nπ±π2

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