$\frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} = z$
${\mathop{\rm Im}\nolimits} (z) = 0$$\theta = ?$
$z = \frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} \times \frac{{1 - i\cos \theta }}{{1 - i\cos \theta }}$
$ = \frac{{1 - i(\cos \theta + \sin \theta ) + {i^2}\sin \theta \cos \theta }}{{1 - {i^2}{{\cos }^2}\theta }}$
$ = \frac{{1 - \sin \theta \cos \theta - i(\cos \theta + \sin \theta )}}{{1 + {{\cos }^2}\theta }}$
$\because {\mathop{\rm Im}\nolimits} (z) = 0,\cos \theta + \sin \theta = 0$
$\Rightarrow \tan \theta = -1 $
$\Rightarrow \theta = n\pi - \frac {\pi}{4}$,$n\in I$