Question-25 The equation \sec ^{2} \theta=\frac{4 x y}{(x+y)^{2}} is possible only when
A) x = y
B) x < y
C) x > y
D) None of these
Answer: A) x = y
Explanation
We have, \sec ^{2} \theta=\frac{4 x y}{(x+y)^{2}}
\because \sec ^{2} \theta \geq 1Then, \frac{4 x y}{(x+y)^{2}} \geq 1
\Rightarrow 4 x y \geq(x+y)^{2} \Rightarrow 4 x y \geq x^{2}+y^{2}+2 x y \Rightarrow(x-y)^{2} \leq 0 \Rightarrow x \leq y\therefore it is possible when x=y.