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

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(D) We know that the range of $\sin^2 	heta$ is $0 \le \sin^2 	heta \le 1$.Given the equation $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,it must sat

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None of these

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(D) We know that the range of $\sin^2 	heta$ is $0 \le \sin^2 	heta \le 1$.Given the equation $\sin^2 	heta = \frac{x^2 + y^2 + 1}{2x}$,it must satisfy the condition $\frac{x^2 + y^2 + 1}{2x} \le 1$.Assuming $x > 0$,we have $x^2 + y^2 + 1 \le 2x$.Rearranging the terms,we get $x^2 - 2x + 1 + y^2 \le 0$.This simplifies to $(x - 1)^2 + y^2 \le 0$.Since the sum of squares of real numbers can only be non-negative,the only way for this inequality to hold is if $(x - 1)^2 = 0$ and $y^2 = 0$.This implies $x = 1$ and $y = 0$.Since the value of $x$ depends on the value of $y$ (i.e.,$x$ is not fixed as $1$ unless $y=0$),and the options provided do not explicitly state this dependency,the correct choice is "None of these".

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

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