The expression [sin (x/2)+cos (x/2)-i tan (x)]/[(1+2isin(x/2)] is real then the set of all possible values of x is (1) n pi+alpha (2) 2n pi (3)n pi/2+alpha (4) None of these

[sin (x/2)+cos (x/2)-i tan (x)]/[(1+2isin(x/2)] 

Multiply numerator and denominator (1-2i sin (x/2))

Since it’s real, equate the imaginary part to 0.

-tan (x/2)-2 sin (x/2)(sin x/2 + cos x/2) = 0

sin (x/2)(2+sin x/2 + cos x/2) = 0

2+sin x/2 + cos x/2 cannot be zero.

sin (x/2) = 0

x/2 = nπ

x = 2nπ

Hence option (2) is the answer.

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