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sin2θ=(x+y)24xy, where x,yR, gives real θ if and only if

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a
x+y=0
b
x=y
c
|x|=|y|0
d
none of these

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detailed solution

Correct option is B

 

We know that

0sin2θ10(x+y)24xy10(x+y)24xy    [4xy>0](xy)20x=y

Thus, we have x=y and xy>0

 |x|=|y|0

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