Inverse Trigonometric Functions Question 130
Question: If $ {{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha , $ then $ 4x^{2}-4xy\cos \alpha +y^{2} $ is equal to
Options:
A) 4
B) $ 2{{\sin }^{2}}\alpha $
C) $ -4{{\sin }^{2}}\alpha $
D) $ 4{{\sin }^{2}}\alpha $
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Answer:
Correct Answer: D
we have $ {{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha $
Or $ x=\cos ( {{\cos }^{-1}}\frac{y}{2}+\alpha ) $
$ =\cos ( {{\cos }^{-1}}\frac{y}{2} )\cos \alpha -\sin ( {{\cos }^{-1}}\frac{y}{2} )\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 $ 4x^{2}+y^{2}{{\cos }^{2}}\alpha -4xy\cos \alpha $
$ =4{{\sin }^{2}}\alpha -y^{2}{{\sin }^{2}}\alpha $
Or $ 4x^{2}-4xy\cos \alpha +y^{2}=4{{\sin }^{2}}\alpha $
BETA
