Question

The real values of $\theta$ for which the expression

$\frac{1-i\sin \theta }{1+2i\sin \theta }$ is purely real is for $n\in Z$,

• A. $n\pi$
• B. $n\frac{\pi }{3}$
• C. $(2n+1)\frac{\pi }{2}$
• D. $(n+1)\frac{\pi }{2}$

Answer 1

The correct option is A $n\pi$

We have,

$\frac{1-isin\theta }{1+2isin\theta }-\left(\frac{\overline{1-isin\theta }}{1+2isin\theta }\right)=0$
$\Rightarrow (1-isin\theta )(1-2isin\theta )-(1+isin\theta )(1+2isin\theta )=0$
$\Rightarrow 1-3isin\theta -2si{n}^2\theta -(1+3isin\theta -2si{n}^2\theta )=0$
$\Rightarrow -6isin\theta =0$
$\Rightarrow sin\theta =0$
$\Rightarrow \theta =n\pi$
Hence, option A is correct