If sin^2 theta = frac{x^2 + y^2 + 1}{2x},then | vedclass

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(D) We know that for any real $	heta$,$0 \le \sin^2 	heta \le 1$.Given $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,we must have $\frac{x^2 + y^2 + 1}

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None of these

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(D) We know that for any real $	heta$,$0 \le \sin^2 	heta \le 1$.Given $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,we must have $\frac{x^2 + y^2 + 1}{2x} \le 1$.Assuming $x > 0$,we get $x^2 + y^2 + 1 \le 2x$,which simplifies to $(x - 1)^2 + y^2 \le 0$.Since squares of real numbers are non-negative,this inequality holds only if $(x - 1)^2 = 0$ and $y^2 = 0$,which implies $x = 1$ and $y = 0$.If $x Since the value of $x$ depends on $y$ (specifically $y$ must be $0$ for $x=1$),$x$ is not uniquely determined as a constant independent of $y$ in all cases,or rather,the condition requires $y=0$ and $x=1$. Given the options,$x=1$ is a specific case,but since $y$ is not defined,the most accurate choice is $D$.

If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,then $x$ must be

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