Choose the correct answer. The real value of for which the expres… - Vidyadip

MCQ

Choose the correct answer.
The real value of $\alpha$ for which the expression $\frac{1-\text{i}\sin\alpha}{1+2\text{i}\sin\alpha}$ is purely real is:

A
$(\text{n}+1)\frac{\pi}{2}$
B
$(2\text{n}+1)\frac{\pi}{2}$
✓
$\text{n}\pi$
D
None of these, where $\text{n}\in\text{N}$

✓

Answer

Correct option: C.

$\text{n}\pi$

$=\frac{(1-\text{i}\sin\alpha)(1-2\text{i}\sin\alpha)}{(1+2\text{i}\sin\alpha)(1-2\text{i}\sin\alpha)}$
$=\frac{1-\text{i}\sin\alpha-2\text{i}\sin\alpha+2\text{i}^2\sin^2\alpha}{1-4\text{i}^2\sin^2\alpha}$
$=\frac{1-3\text{i}\sin\alpha-2\sin^2\alpha}{1+4\sin^2\alpha}$
$=\frac{1-2\sin^2\alpha}{1+4\sin^2\alpha}-\frac{3\text{i}\sin\alpha}{1+4\sin^2\alpha}$
It is given that $z$ is a purely real.
$\Rightarrow\frac{-3\text{i}\sin\alpha}{1+4\sin^2\alpha}=0$
$\Rightarrow-3\sin\alpha=0$
$\Rightarrow\sin\alpha=0$
$\Rightarrow\alpha=\text{n}\pi,\text{ n}\in\text{I}$

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