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 :

• A. $\pi -{\tan }^{-1}\left(\frac{4}{3}\right)$
• B. $-{\tan }^{-1}\left(\frac{3}{4}\right)$
• C. $\pi -{\tan }^{-1}\left(\frac{3}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{4}{3}\right)$

Answer 1

The correct option is A $\pi -{\tan }^{-1}\left(\frac{4}{3}\right)$

Let $z=\frac{3+i\sin \theta }{4-i\cos \theta }\times \frac{4+i\cos \theta }{4+i\cos \theta }$

$=\frac{12-\sin \theta \cos \theta +i(4\sin \theta +3\cos \theta )}{16+{\cos }^2\theta }$
Given, $z$ is real
$\therefore 4\sin \theta +3\cos \theta =0$
$\Rightarrow \tan \theta =-\frac{3}{4}$
$\Rightarrow \theta$ lies in $2^{nd}$ quadrant
or, $\theta$ lies in $4^{th}$ quadrant.
$\therefore \arg (\sin \theta +i\cos \theta )=\{\begin{array}{c}\pi +{\tan }^{-1}\left(\frac{\cos \theta }{\sin \theta }\right),\theta \in (\frac{\pi }{2},\pi )\\ 2\pi +{\tan }^{-1}\left(\frac{\cos \theta }{\sin \theta }\right),\theta \in (\frac{3\pi }{2},2\pi )\end{array}$

$\Rightarrow \arg (\sin \theta +i\cos \theta )=\{\begin{array}{c}\pi -{\tan }^{-1}\left(\frac{4}{3}\right),\theta \in (\frac{\pi }{2},\pi )\\ 2\pi -{\tan }^{-1}\left(\frac{4}{3}\right),\theta \in (\frac{3\pi }{2},2\pi )\end{array}$