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Question Number 37745 by kunal1234523 last updated on 17/Jun/18

$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{equation}{\sec }^2\theta =\frac{4xy}{{(x+y)}^2}\mathrm{is}$$$$\mathrm{only}\mathrm{possible}\mathrm{when}x=y$$

Answered by math1967 last updated on 17/Jun/18

$$x,yreal\therefore {(x-y)}^2⩾0$$$${(x+y)}^2-4xy⩾0\Rightarrow {(x+y)}^2⩾4xy$$$$\Rightarrow 4xy⩽{(x+y)}^2$$$$\Rightarrow \frac{4xy}{{(x+y)}^2}⩽1but{sec}^2\theta ⩾1$$$$nowifx=ythen\frac{4xy}{{(x+y)}^2}=1$$$$\therefore {\sec }^2\theta =\frac{4xy}{{(x+y)}^2}ispossiblewhenx=y$$

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