Question

A value of $\theta$ such that $\frac{2+i\sin \theta }{1-2i\sin \theta }$ is purely real is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\pi$

Answer 1

The correct option is D $\pi$

$\frac{2+i\sin \theta }{1-2i\sin \theta }=\frac{2+i\sin \theta }{1-2i\sin \theta }\times \frac{1+2i\sin \theta }{1+2i\sin \theta }=\frac{1+5i\sin \theta -2{\sin }^2\theta }{1+4{\sin }^2\theta }=\frac{1-2{\sin }^2\theta }{1+4{\sin }^2\theta }+\frac{5i\sin \theta }{1+4{\sin }^2\theta }$
So,for it to be purely real
$\frac{5\sin \theta }{1+4{\sin }^2\theta }=0\Rightarrow \sin \theta =0\Rightarrow \theta \in n\pi ,n\in Z$