A) \[3x-4\]
B) \[4x-3\]
C) \[4x+3\]
D) None of these
Correct Answer: B
Solution :
[b]: Consider,\[x+iy=\frac{3}{2+\cos \theta +isin\theta }\] \[=\frac{3(2+cos\theta -i\,sin\theta )}{{{(2+\cos \theta )}^{2}}+si{{n}^{2}}\theta }\] \[=\frac{6+3cos\theta -3i\,sin\theta }{4+{{\cos }^{2}}\theta +4\cos \theta +{{\sin }^{2}}\theta }\] \[=\left[ \frac{6+3cos\theta }{5+4\cos \theta } \right]+i\left[ \frac{-3\sin \theta }{5+4\cos \theta } \right]\] \[\Rightarrow \]\[x=\frac{3(2+cos\theta )}{5+4\cos \theta },y=\frac{-3\sin \theta }{5+4\cos \theta }\] \[\therefore \]\[{{x}^{2}}+{{y}^{2}}=\frac{9[4+co{{s}^{2}}+4cos+si{{n}^{2}}]}{{{(5+4cos\theta )}^{2}}}\] \[=\frac{9}{5+4cos\theta }=4\left[ \frac{6+3\cos \theta }{5+4\cos \theta } \right]-3=4x-3\]
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