Inverse Trigonometric Functions Question 37
Question: If $ {{\cos }^{-1}}x-{{\cos }^{-1}}\frac{y}{2}=\alpha $ , then $ 4x^{2}-4xy\cos \alpha +y^{2} $ is equal to
[AIEEE 2005]
Options:
A) $ 4{{\sin }^{2}}\alpha $
B) $ -4{{\sin }^{2}}\alpha $
C) $ 2\sin 2\alpha $
D) $ 4 $
Show Answer
Answer:
Correct Answer: A
Solution:
If $ {{\cos }^{-1}}\frac{x}{a}+{{\cos }^{-1}}\frac{y}{b}=\theta $ Then $ \frac{x^{2}}{a^{2}}\cos \theta +\frac{y^{2}}{b^{2}}={{\sin }^{2}}\theta $ Here $ {{\cos }^{-1}}\frac{x}{1}+{{\cos }^{-1}}\frac{y}{2}=\alpha ; $
$ \therefore \frac{x^{2}}{1}-\frac{2xy}{2}\cos \alpha +\frac{y^{2}}{4}={{\sin }^{2}}\alpha $
$ x^{2}-xy\cos \alpha +\frac{y^{2}}{4}={{\sin }^{2}}\alpha $
$ 4x^{2}-4xy\cos \alpha +y^{2}=4{{\sin }^{2}}\alpha $ .
BETA
