• # question_answer If $x+iy=\frac{3}{2+\cos \theta +i\sin \theta },$then ${{x}^{2}}+{{y}^{2}}$ is equal to A) $3x-4$ B) $4x-3$ C) $4x+3$ D) None of these

If $x+iy=\frac{3}{2+\cos \theta +i\sin \theta }$ $=\frac{3(2+\cos \theta -i\sin \theta )}{{{(2+\cos \theta )}^{2}}+{{\sin }^{2}}\theta }=\frac{6+3\cos \theta -3i\sin \theta }{4+{{\cos }^{2}}\theta +4\cos \theta +{{\sin }^{2}}\theta }$ $=\left[ \frac{6+3\cos \theta }{5+4\cos \theta } \right]+i\,\left[ \frac{-3\sin \theta }{5+4\cos \theta } \right]$ Þ  $x=\frac{3(2+\cos \theta )}{5+4\cos \theta },y=\frac{-3\sin \theta }{5+4\cos \theta }$ \ ${{x}^{2}}+{{y}^{2}}=\frac{9}{{{(5+4\cos \theta )}^{2}}}$$[4+{{\cos }^{2}}\theta +4\cos \theta +{{\sin }^{2}}\theta ]$ $=\frac{9}{5+4\cos \theta }=4\left[ \frac{6+3\cos \theta }{5+4\cos \theta } \right]-3=4x-3$ Trick: $x+iy=\frac{3(2+\cos \theta -i\sin \theta )}{(2+\cos \theta +i\sin \theta )(2+\cos \theta -i\sin \theta )}$ Let$y=0$, then$\sin \theta =0$i.e.,$\theta =0$ Now put $x=1$ then${{x}^{2}}+{{y}^{2}}={{1}^{2}}+0=1$. Also option b gives 4(1) - 3=1.