The real value of θ for which the expression 1−isin⁡θ1+2isin⁡θ is purely real is

z=(1−isin⁡θ)(1−2isin⁡θ)1+4sin2⁡θ=1−2sin2⁡θ−3isin⁡θ1+4sin2⁡θIt will be purely real if I.P. =0, i.e. sin⁡θ=0 ∴θ=nπ

The real value of θ for which the expression 1−isin⁡θ1+2isin⁡θ is purely real is

z=(1−isin⁡θ)(1−2isin⁡θ)1+4sin2⁡θ=1−2sin2⁡θ−3isin⁡θ1+4sin2⁡θIt will be purely real if I.P. =0, i.e. sin⁡θ=0 ∴θ=nπ

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The real value of $θ$ for which the expression $\frac{1 - i \sin θ}{1 + 2 i \sin θ}$ is purely real is

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$n π$

$(n + 1) π / 2$

$(2 n + 1) π / 2$

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answer is A.

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$\begin{array}{l} z = \frac{(1 - i \sin θ) (1 - 2 i \sin θ)}{1 + 4 \sin^{2} θ} \\ = \frac{(1 - 2 \sin^{2} θ) - 3 i \sin θ}{1 + 4 \sin^{2} θ} \end{array}$

It will be purely real if $I.P. = 0, i.e. \sin θ = 0 ∴ θ = n π$

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The real value of θ for which the expression 1−isin⁡θ1+2isin⁡θ is purely real is

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The real value of $θ$ for which the expression $\frac{1 - i \sin θ}{1 + 2 i \sin θ}$ is purely real is