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

$z=\frac{\left(3+i\sin \theta \right)}{\left(4-i\cos \theta \right)}\times \frac{\left(4+i\cos \theta \right)}{4+i\cos \theta }$

as z is purely real $\Rightarrow 3\cos \theta +4\sin \theta =0\Rightarrow \tan \theta =-\frac{3}{4}$

$$\begin{gathered} \therefore \arg \left(\sin \theta +i\cos \theta \right)=\pi -{\tan }^{-1}\left|\frac{\cos \theta }{\sin \theta }\right| or -{\tan }^{-1}\left|\frac{\cos \theta }{\sin \theta }\right| \\ =\pi -{\tan }^{-1}\left(\frac{4}{3}\right) or -{\tan }^{-1}\left(\frac{4}{3}\right) \end{gathered}$$