A) 1
B) 2
C) 4
D) 9
Correct Answer: A
Solution :
| [a] \[x{{\sin }^{3}}\alpha +y{{\cos }^{3}}\alpha =\sin \alpha cos\alpha \] \[\Rightarrow \]\[x\sin \alpha \cdot {{\sin }^{2}}\alpha +y\cos \alpha \cdot {{\cos }^{2}}\alpha \] \[=\sin \alpha \cos \alpha \] \[\Rightarrow \]\[x\sin \alpha .{{\sin }^{2}}\alpha +x\sin \alpha .{{\cos }^{2}}\alpha \] \[=\sin \alpha \cos \alpha \]\[[\because y\,cos\alpha =x\,sin\,\alpha ]\] \[\Rightarrow \]\[x\sin \alpha (si{{n}^{2}}\alpha +co{{s}^{2}}\alpha )=\sin \alpha \cos \alpha \] \[\therefore \] \[x=\cos \alpha \] And \[y\cos \alpha =cos\alpha sin\alpha \]\[\Rightarrow \]\[y=\sin \alpha \] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\] |
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