question_answer

If\[{{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha ,\]then\[4{{x}^{2}}-4xy\,\cos \alpha +{{y}^{2}}\] is equal to     AIEEE  Solved  Paper-2005

A) \[-4\text{ }{{\sin }^{2}}\alpha \]     

B)        \[4\text{ }{{\sin }^{2}}\alpha \] 

C)        4                             

D)        \[2\text{ }{{\sin }^{2}}\alpha \]

Correct Answer: B

Solution :

Given that On squaring both sides, we get                                                         \[\Rightarrow \,\,4{{x}^{2}}\,-4\,xy\,\cos \alpha \,={{y}^{2}}\,=4{{\sin }^{2}}\,\alpha \]