Question:
If $\frac{3+i \sin \theta}{4-i \cos \theta}, \theta \in[0,2 \pi]$, is $a$ real number, then an argument of $\sin \theta+i \cos \theta$ is:
Correct Option: , 2
Solution:
Let $z=\frac{3+i \sin \theta}{4-i \cos \theta}$, after rationallsing
$z=\frac{(3+i \sin \theta)}{(4-i \cos \theta)} \times \frac{(4+i \cos \theta)}{(4+i \cos \theta)}$
As $z$ is purely real
$\Rightarrow 3 \cos \theta+4 \sin \theta=0 \Rightarrow \tan \theta=-\frac{3}{4}$
$\arg (\sin \theta+i \cos \theta)=\pi+\tan ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right)$
$=\pi+\tan ^{-1}\left(-\frac{4}{3}\right)=\pi-\tan ^{-1}\left(\frac{4}{3}\right)$
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