Question

$\sin ^{2} \theta=\frac{4 x y}{(x+y)^{2}}$ is true if and only if :

• A. $x+y \neq 0$
• B. $x=y, x \neq 0, y \neq 0$
• C. $x = y$
• D. $x \neq 0, y \neq 0$

Answer 1

Answer: (A)

$\because \sin \theta \leq 1 $
$ \sin ^{2} \theta \leq 1 $
$ \Rightarrow \frac{4 x y}{(x+y)^{2}} \leq 1 $
$ \Rightarrow 0 \leq(x+y)^{2}-4 x y $
$ \Rightarrow x^{2}+y^{2}+2 x y-4 x y \geq 0 $
$ \Rightarrow (x-y)^{2} \geq 0 $
Which is true for all real $x$ and $y$ provided $x+y \neq 0,$ otherwise $\frac{4 x y}{(x+y)^{2}}$ will be meaningless.
Note $: \sec ^{2} \theta \leq \frac{4 x y}{(x+y)^{2}}$ is possible only, if $x=y$.