Question

If $\frac{3+i\sin \theta }{4-i\cos \theta },\theta \in [0,2\pi ]$ is a real number, then argument of $\sin \theta +i\cos \theta$ can be:

• A. $\pi -\alpha ,$ where $\tan \alpha =\frac{4}{3}$
• B. $-\alpha ,$ where $\tan \alpha =\frac{3}{4}$
• C. $\pi -\alpha ,$ where $\tan \alpha =\frac{3}{4}$
• D. $-\alpha ,$ where $\tan \alpha =\frac{4}{3}$

Answer 1

Answer: (A), (D)

$-\alpha ,$ where $\tan \alpha =\frac{4}{3}$

Let $Z=\frac{3+i\sin \theta }{4-i\cos \theta }$
Now,
$Z=\frac{3+i\sin \theta }{4-i\cos \theta }\times \frac{4+i\cos \theta }{4+i\cos \theta }\Rightarrow Z=\frac{12-\sin \theta \cos \theta +i(4\sin \theta +3\cos \theta )}{16+{\cos }^2\theta }$
Given, $Z$ is real
$\therefore 4\sin \theta +3\cos \theta =0\Rightarrow \tan \theta =-\frac{3}{4}$

$\text{Case-1:}$ If $\theta$ lies in $2^{nd}$ quadrant
$z=\sin \theta +i\cos \theta$
$Re\left(z\right)>0\&Im\left(z\right)<0$
So, $z$ lies in $4^{th}$ quadrant
$\tan \alpha =\frac{|\cos \theta |}{|\sin \theta |}=-\cot \theta =\frac{4}{3}$
$\therefore \arg \left(z\right)=-\alpha ,$ where $\tan \alpha =\frac{4}{3}$

$\text{Case-2:}$ If $\theta$ lies in $4^{th}$ quadrant
$z=\sin \theta +i\cos \theta$
$Re\left(z\right)<0\&Im\left(z\right)>0$
So, $z$ lies in $2^{nd}$ quadrant
$\tan \alpha =\frac{|\cos \theta |}{|\sin \theta |}=\frac{\cos \theta }{-\sin \theta }=-\cot \theta =\frac{4}{3}$
$\therefore \arg \left(z\right)=\pi -\alpha ,$ where $\tan \alpha =\frac{4}{3}$