A) 4
B) \[2{{\sin }^{2}}\alpha \]
C) \[-4{{\sin }^{2}}\alpha \]
D) \[4{{\sin }^{2}}\alpha \]
Correct Answer: D
Solution :
[d] we have \[{{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha \] Or \[x=\cos \left( {{\cos }^{-1}}\frac{y}{2}+\alpha \right)\] \[=\cos \left( {{\cos }^{-1}}\frac{y}{2} \right)\cos \alpha -\sin \left( {{\cos }^{-1}}\frac{y}{2} \right)\sin \alpha \] \[=\frac{y}{2}\cos \alpha -\sqrt{1-\frac{{{y}^{2}}}{4}}\sin \alpha \] Or \[2x=y\cos \alpha -\sin \alpha \sqrt{4-{{y}^{2}}}\] Or \[2x-y\cos \alpha =-\sin \alpha \sqrt{4-{{y}^{2}}}\] Squaring, we get \[4{{x}^{2}}+{{y}^{2}}{{\cos }^{2}}\alpha -4xy\cos \alpha \] \[=4{{\sin }^{2}}\alpha -{{y}^{2}}{{\sin }^{2}}\alpha \] Or \[4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha \]
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