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

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Question

(B) We know that for any real numbers $x$ and $y$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{x+y}{2} \geq \sqrt{xy

Vedclass · Accepted Answer

$x = y, x 
eq 0$

Vedclass · Answer

(B) We know that for any real numbers $x$ and $y$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{x+y}{2} \geq \sqrt{xy}$.Squaring both sides,we get $\frac{(x+y)^2}{4} \geq xy$,which implies $\frac{4xy}{(x+y)^2} \leq 1$.Since $cosec^2 	heta \geq 1$ for all real $	heta$,the equation $cosec^2 	heta = \frac{4xy}{(x+y)^2}$ can only hold if both sides are equal to $1$.This requires $\frac{4xy}{(x+y)^2} = 1$,which simplifies to $4xy = (x+y)^2$,or $(x-y)^2 = 0$,meaning $x = y$.Additionally,since $cosec^2 	heta$ is undefined if the denominator is zero,we must have $x+y 
eq 0$. Given $x=y$,this implies $2x 
eq 0$,so $x 
eq 0$.Thus,the condition is $x = y$ and $x 
eq 0$.

$cosec^2 \theta = \frac{4xy}{(x + y)^2}$ is true if and only if

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