Question

If $cose{c}^2\theta =\frac{4xy}{(x+y{)}^2}$, then

• A. $x=-y$
• B. $x=\frac{1}{y}$
• C. $x=\frac{1}{x}$
• D. $x=y$

Answer 1

The correct option is D $x=y$

$cose{c}^2\theta =\frac{4xy}{(x+y{)}^2}\ge 1\Rightarrow (x+y{)}^2\le 4xy\Rightarrow (x-y{)}^2\le 0\Rightarrow (x-y{)}^2=0\Rightarrow x=y.$