Inverse Trigonometric Functions Question 138
Question: If $ {{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha $ then $ 4x^{2}-2xy\cos \alpha +y^{2} $ is equal to
Options:
A) 2 sin $ \alpha $
B) 4
C) $ 4{{\sin }^{2}}\alpha $
D) $ -4{{\sin }^{2}}\alpha $
Show Answer
Answer:
Correct Answer: C
$ {{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha $
$ \Rightarrow {{\cos }^{-1}}( \frac{xy}{2}+\sqrt{(1-x^{2})( 1-\frac{y^{2}}{4} )} )=\alpha $
$ \Rightarrow {{\cos }^{-1}}( \frac{xy+\sqrt{4-y^{2}-4x^{2}+x^{2}y^{2}}}{2} )=\alpha $
$ \Rightarrow \sqrt{4-y^{2}-4x^{2}+x^{2}y^{2}}=2\cos \alpha -xy $
$ \Rightarrow 4-y^{2}-4x^{2}+x^{2}y^{2}=4{{\cos }^{2}}\alpha $ $ +x^{2}y^{2}-4xy\cos \alpha $
$ \Rightarrow $ $ 4x^{2}+y^{2}-4xy\cos \alpha =4{{\sin }^{2}}\alpha $
BETA
