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

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Question

(d) ${\sin ^2}	heta \le 1$ 

$	herefore$ $\frac{{{x^2} + {y^2} + 1}}{{2x}} \le 1$

$\Rightarrow$ ${x^2} + {y^2} - 2x + 1 \le 0$.

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(d) ${\sin ^2}	heta \le 1$ 

$	herefore$ $\frac{{{x^2} + {y^2} + 1}}{{2x}} \le 1$

$\Rightarrow$ ${x^2} + {y^2} - 2x + 1 \le 0$.

$\Rightarrow$ ${(x - 1)^2} + {y^2} \le 0$

It is possible, if $x = 1$ and $y = 0$, 

$i.e.$, It also depends on value of $y$.

Hence, option $(d)$ is correct.

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

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