The real value of ' α ' for which (1-i sin α)/(1+2 i sin α) is purely real is

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) n π, n ∈ N. Given, (1-i sin α)/(1+2^j sin α) is purely real i.e. (1-i sin α)/(1+2 i sin α) × ((1-2 i sin α))/((1-2 i sin α)) =(1-i sin α-2 i sin α+2 i^2 sin^2 α)/(1-4 i^2 sin^2 α) =(1-3 i sin α-2 sin^2 α)/(1+4 sin^2 α) =(1-2 sin^2 α)/(1+4 sin^2 α)+i ((-3 sin α)/(1+4 sin^2 α) ) which is given to purely real ⇒ (-3 sin α)/(1+4 sin^2 α)=0 ⇒ -3 sin α=0 ⇒ sin α=0 Hence, α=n π, n ∈ N

The real value of ' $\alpha$ ' for which $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is

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