Trigonometric Identities Question 232
Question: For which real values of x and y, the equation $ {{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}} $ is possible ?
Options:
A) $ x=y $
B) $ x>y $
C) $ x<y $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation, $ {{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}} $ Since range of $ \sec \theta $ is $ (-\infty ,-1],\cup [1,\infty ). $
$ \therefore {{\sec }^{2}}\theta \ge 1 $
$ \Rightarrow \frac{4xy}{{{(x+y)}^{2}}}\ge 1 $
$ \Rightarrow {{(x-y)}^{2}}\le 0 $ But $ {{(x-y)}^{2}}</0 $ for any x, $ y\in R $
$ \therefore ,{{(x-y)}^{2}}=0\Rightarrow x=y $
BETA
