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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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x+y=0

x=y

|x|=|y|≠0

none of these

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detailed solution

Correct option is B

We know that0≤sin2⁡θ≤1⇒0≤(x+y)24xy≤1⇒0≤(x+y)2≤4xy [∵4xy>0]⇒(x−y)2≤0⇒x=yThus, we have x=y and xy>0∴ |x|=|y|≠0

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sin2⁡θ=(x+y)24xy, where x,y∈R, gives real θ if and only if

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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