If operatorname{cosec}^{2} theta = frac{4xy}{ | vedclass

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Question

(C) We know that for any real angle $	heta$,the value of $\operatorname{cosec}^{2} 	heta \geq 1$.Given $\operatorname{cosec}^{2} 	heta = \frac{4xy}

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$x = y$

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(C) We know that for any real angle $	heta$,the value of $\operatorname{cosec}^{2} 	heta \geq 1$.Given $\operatorname{cosec}^{2} 	heta = \frac{4xy}{(x+y)^{2}}$,therefore $\frac{4xy}{(x+y)^{2}} \geq 1$.This implies $4xy \geq (x+y)^{2}$ (assuming $(x+y)^2 > 0$).Rearranging the terms,we get $4xy - (x^{2} + 2xy + y^{2}) \geq 0$.This simplifies to $-(x^{2} - 2xy + y^{2}) \geq 0$,which is $-(x-y)^{2} \geq 0$.Multiplying by $-1$,we get $(x-y)^{2} \leq 0$.Since the square of any real number cannot be negative,the only possibility is $(x-y)^{2} = 0$.Therefore,$x - y = 0$,which means $x = y$.

If $\operatorname{cosec}^{2} \theta = \frac{4xy}{(x+y)^{2}},$ then

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