If sin theta = frac{1}{2} left( sqrt{frac{x}{ | vedclass

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(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean ($AM$ $\ge$ $GM$).Applyi

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$x = y$

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(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean ($AM$ $\ge$ $GM$).Applying this to $\sqrt{\frac{x}{y}}$ and $\sqrt{\frac{y}{x}}$,we get:$\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight) \ge \sqrt{\sqrt{\frac{x}{y}} \cdot \sqrt{\frac{y}{x}}}$$\Rightarrow \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight) \ge \sqrt{1} = 1$Since $\sin 	heta$ cannot exceed $1$,the expression must be equal to $1$.$\sin 	heta = 1 \Rightarrow \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight) = 1$This equality holds if and only if $\sqrt{\frac{x}{y}} = \sqrt{\frac{y}{x}}$,which implies $\frac{x}{y} = 1$,or $x = y$.

If $\sin \theta = \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right)$,where $x, y \in R - \{0\}$. Then:

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