Question : The equation $\cos ^{2}\theta=\frac{(x+y)^{2}}{4xy}$ is only possible when,
Option 1: $x=-y$
Option 2: $x>y$
Option 3: $x=y$
Option 4: $x<y$
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Correct Answer: $x=y$
Solution :
$\cos ^{2}\theta=\frac{(x+y)^{2}}{4xy}$
We know that $(x+y)\geq 2\sqrt{xy}$ (Since the arithmetic mean is always greater than or equal to the geometric mean)
$⇒(x+y)^2\geq 4xy$
$⇒\frac{(x+y)^2}{4xy}\geq 1$
Since $0\leq \cos ^{2}\theta\leq 1$, solution exists when $\frac{(x+y)^2}{4xy}=1$
$\therefore \frac{(x+y)^2}{4xy}=1$
$⇒4xy =x^2+y^2 + 2xy$
$⇒x^2+y^2 - 2xy=0$
$⇒(x-y)^2 = 0$
$⇒x-y=0$
$\therefore x=y$
Hence, the correct answer is $x=y$.
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