A) - 1
B) ±1
C) 1
D) None of these
Correct Answer: C
Solution :
We have \[x\,{{\sin }^{3}}\alpha +y\,{{\cos }^{3}}\alpha =\sin \,\alpha \,\cos \,\alpha \] ?..(i) and \[x\,\sin \,\alpha -y\,\cos \,\alpha =0\] ?..(ii) Now from (ii), \[x\,\sin \,\alpha =y\,\cos \,\alpha \] Putting in (i), we get \[\Rightarrow \,\,y\,\cos \alpha \,{{\sin }^{2}}\alpha +y\,{{\cos }^{3}}\alpha =\sin \,\alpha \,\cos \,\alpha \] \[\Rightarrow \,\,y\,\cos \alpha \,\left\{ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right\}=\sin \,\alpha \,\cos \,\alpha \] \[\Rightarrow \,\,y\,\cos \,\alpha =\sin \,\alpha \,\cos \,\alpha \,\Rightarrow \,\,y=\sin \,\alpha \] and \[x=\cos \,\alpha \] Hence, \[{{x}^{2}}+{{y}^{2}}={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1.\]
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