If the equation x^{2}+y^{2}-2xy \sin^{2}\theta=0 contains <strong>REAL</strong> solution for x and y, then

Explanation

The equation is (x-y)^2 + 2xy - 2xy\sin^2\theta = 0 \implies (x-y)^2 + 2xy\cos^2\theta = 0. For real x and y, both terms must be 0, which implies x=y and either x=0 or \cos\theta=0. Treating as quadratic in x, discriminant \geq 0 \implies 4y^2\sin^4\theta - 4y^2 \geq 0 \implies y^2(\sin^4\theta - 1) \geq 0. Thus \sin^2\theta = 1, making x^2+y^2-2xy=0, so x=y.

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