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

Let z = ${{3 + i\sin \theta } \over {4 - i\cos \theta }}$ <br><br>= $${{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}$$ <br><br>= $${{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}$$ <br><br>As Z is purely real <br><br>$\therefore$ 4sin$\theta$ + 3cos$\theta$ = 0 <br><br>$\Rightarrow$ tan $\theta$ = $- {3 \over 4}$ <br><br>$\therefore$ $\theta$ lies in the 2^nd quadrant then <br><br>arg(sinθ + icosθ) = $\pi$ + ${\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)$ <br><br>= $\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)$