Trigonometric Identities Question 165
Question: If $ 2y,\cos \theta =x\sin ,\theta \text{ and }2x\sec \theta -y,cosec,\theta =3, $ then $ x^{2}+4y^{2}= $
[WB JEE 1988]
Options:
A) 4
B) - 4
C) ± 4
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ 2y\cos \theta =x,\sin \theta $ ?..(i) and $ 2x,\sec \theta -ycosec,\theta =3 $ ?..(ii)
$ \Rightarrow \frac{2x}{\cos \theta }-\frac{y}{\sin \theta }=3 $
$ \Rightarrow 2x,\sin \theta -y,\cos \theta -3,\sin \theta \cos \theta =0 $ ?..(iii) Solving (i) and (iii), we get $ y=\sin \theta $ and $ x=2\cos \theta $ Now, $ x^{2}+4y^{2}=4{{\cos }^{2}}\theta +4{{\sin }^{2}}\theta $ $ =4,({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=4 $ .
BETA
