sin^{2} theta = frac{(x+y)^{2}}{4xy} is possi | vedclass

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(B) We know that for any real $	heta$,the range of $\sin 	heta$ is $[-1, 1]$.Therefore,$\sin^{2} 	heta$ must satisfy $0 \leq \sin^{2} 	heta \leq 1

Vedclass · Accepted Answer

$x > 0, y > 0, x = y$

Vedclass · Answer

(B) We know that for any real $	heta$,the range of $\sin 	heta$ is $[-1, 1]$.Therefore,$\sin^{2} 	heta$ must satisfy $0 \leq \sin^{2} 	heta \leq 1$.Given $\sin^{2} 	heta = \frac{(x+y)^{2}}{4xy}$,we must have $\frac{(x+y)^{2}}{4xy} \leq 1$.Since $(x+y)^{2} \geq 0$,for the expression to be defined and positive,$xy > 0$ (meaning $x$ and $y$ must have the same sign).If $xy > 0$,then $(x+y)^{2} \leq 4xy$.$(x+y)^{2} - 4xy \leq 0$.$(x-y)^{2} \leq 0$.Since the square of a real number cannot be negative,$(x-y)^{2} \leq 0$ is only possible if $(x-y)^{2} = 0$,which implies $x = y$.If $x = y$,then $\sin^{2} 	heta = \frac{(2x)^{2}}{4x^{2}} = \frac{4x^{2}}{4x^{2}} = 1$,which is a valid value for $\sin^{2} 	heta$.

$\sin^{2} \theta = \frac{(x+y)^{2}}{4xy}$ is possible only when

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