A) \[2\text{ }si{{n}^{2}}\alpha \]
B) 4
C) \[\text{4 }si{{n}^{2}}\alpha \]
D) \[\text{-4 }si{{n}^{2}}\alpha \]
Correct Answer: C
Solution :
\[{{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha \] \[\Rightarrow \]\[{{\cos }^{-1}}\left( \frac{xy}{2}+\sqrt{1-{{x}^{2}}}\sqrt{1-\frac{{{y}^{2}}}{4}} \right)=\alpha \] \[\Rightarrow \]\[\frac{xy}{2}+\sqrt{1-{{x}^{2}}}\sqrt{1-\frac{{{y}^{2}}}{4}}=\cos \alpha \] \[\Rightarrow \]\[2\sqrt{1-{{x}^{2}}}\sqrt{\frac{4-{{y}^{2}}}{4}}=2\cos \alpha -xy\] \[\Rightarrow \]\[\frac{4(1-{{x}^{2}})(4-{{y}^{2}})}{4}\] \[=4{{\cos }^{2}}\alpha +{{x}^{2}}{{y}^{2}}-4xy\cos \alpha \] \[\Rightarrow \]\[4-4{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}\] \[=4{{\cos }^{2}}\alpha +{{x}^{2}}{{y}^{2}}-4xy\cos \alpha \] \[\Rightarrow \]\[4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha \]
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