A) \[\sin \theta -\cos \theta \]
B) \[\sin \theta +\cos \theta \]
C) \[0\]
D) \[1\]
Correct Answer: D
Solution :
[d] Given, \[x{{\sin }^{3}}\theta +yy{{\cos }^{3}}\theta =\sin \theta \cos \theta \] (i) and \[x\sin \theta -y\cos \theta =0\] ...(ii) From Eq. (ii), we get \[x\sin \theta =y\cos \theta \] ...(iii) On putting this value in Eq. (i), we get \[y\cos \theta .{{\sin }^{2}}\theta +y{{\cos }^{3}}\theta =\sin \theta \cos \theta \] \[\Rightarrow \] \[y\cos \theta \,(si{{n}^{2}}\theta +co{{s}^{2}}\theta )=\sin \theta \cos \theta \] \[\Rightarrow \] \[y=\sin \theta \] From Eq. (iii), we get \[x=\cos \theta \] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] |
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