The equation {sin ^2}theta = frac{ | vedclass

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Now, $\sin ^{2} 	heta=\frac{x^{2}+y^{2}}{2 x y}$

$	herefore \mathrm{x},$ $y$ have same sign

$\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left[(\sqrt

Vedclass · Accepted Answer

$x = y$

Vedclass · Answer

Now, $\sin ^{2} 	heta=\frac{x^{2}+y^{2}}{2 x y}$

$	herefore \mathrm{x},$ $y$ have same sign

$\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left[(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}})^{2}+2ight] \geq 1$

Now, $\sin ^{2} 	heta \leq 1 .$

Therefore, $\frac{x^{2}+y^{2}}{2 x y}=1$

$  \Rightarrow x=y$

The equation ${\sin ^2}\theta = \frac{{{x^2} + {y^2}}}{{2xy}},x,y, \ne 0$ is possible if

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