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$\sin^{2} θ = \frac{(x + y)^{2}}{4 x y}$ , where $x, y \in R$ , gives real $θ$ if and only if

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By Expert Faculty of Sri Chaitanya

a

x + y = 0

b

x = y

c

| x | = | y | \neq 0

d

none of these

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detailed solution

Correct option is B

We know that

$\begin{array}{l} 0 \leq \sin^{2} θ \leq 1 \\ \Rightarrow 0 \leq \frac{(x + y)^{2}}{4 x y} \leq 1 \\ \Rightarrow 0 \leq (x + y)^{2} \leq 4 x y [∵ 4 x y > 0] \\ \Rightarrow (x - y)^{2} \leq 0 \Rightarrow x = y \end{array}$

Thus, we have $x = y$ and $x y > 0$

$∴ | x | = | y | \neq 0$

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