Trigonometric Identities Question 110
Question: The equation $ {{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}} $ is only possible when
[MP PET 1986; IIT 1996]
Options:
A) $ x=y $
B) $ x<y $
C) $ x>y $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ {{\cos }^{2}}\theta \le 1 $ $ {{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}}\ge 1\Rightarrow 4xy\ge {{(x+y)}^{2}}\Rightarrow {{(x-y)}^{2}}\le 0 $ It is possible only when $ x=y $ , $ (\because x,,y\in R) $ .
BETA
