A) \[0\]
B) \[1/2\]
C) \[1\]
D) \[3/2\]
Correct Answer: C
Solution :
| [c] We have, \[x{{\sin }^{3}}\theta +y{{\cos }^{3}}\theta =\sin \theta \cos \theta \] |
| \[(x\sin \theta ){{\sin }^{2}}\theta +(y\cos \theta ){{\cos }^{2}}\theta =\sin \theta \cos \theta \] |
| \[x\sin \theta ({{\sin }^{2}}\theta )+(x\sin \theta ){{\cos }^{2}}\theta =\sin \theta \cos \theta \] |
| \[x\sin \theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=\sin \theta \cos \theta \] |
| \[x\sin \theta =\sin \theta \cos \theta \,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,x=\cos \theta \] |
| Now,\[x\sin \theta =y\cos \theta \] |
| \[\cos \theta \sin \theta =y\cos \theta \] |
| \[y=\sin \theta \] |
| Hence, \[{{x}^{2}}+{{y}^{2}}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] |
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