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Question Number 215679 by BaliramKumar last updated on 14/Jan/25

Answered by MATHEMATICSAM last updated on 14/Jan/25

$${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}$$$$\Rightarrow {\cos }^2\theta =\frac{{(x+y)}^2}{4xy}$$$${(x-y)}^2⩾0$$$$\Rightarrow {(x+y)}^2-4xy⩾0$$$$\Rightarrow {(x+y)}^2⩾4xy$$$$\Rightarrow \frac{{(x+y)}^2}{4xy}⩾1$$$$\mathrm{and}\mathrm{its}\mathrm{equal}\mathrm{to}1\mathrm{when}x=y$$$$\mathrm{We}\mathrm{know}{\cos }^2\theta ⩽1$$$$\mathrm{So}{\cos }^2\theta \mathrm{will}\mathrm{be}\frac{{(x+y)}^2}{4xy}\mathrm{when}x=y.$$$$\mathrm{So}{\sec }^2\theta =\frac{4xy}{{(x+y)}^2}\mathrm{is}\mathrm{only}\mathrm{possible}\mathrm{when}$$$$x=y.$$

Answered by A5T last updated on 14/Jan/25

$${(\mathrm{x}+\mathrm{y})}^2⩾4\mathrm{xy}\Rightarrow {\sec }^2\theta =\frac{4\mathrm{xy}}{{(\mathrm{x}+\mathrm{y})}^2}⩽\frac{4\mathrm{xy}}{4\mathrm{xy}}=1$$$$\mathrm{But}{\sec }^2\theta =\frac{1}{{\cos }^2\theta }⩾1$$$$\Rightarrow {\sec }^2\theta ⩾1\mathrm{and}{\sec }^2\theta ⩽1\mathrm{which}\mathrm{is}\mathrm{only}\mathrm{possible}$$$$\mathrm{when}\sec \theta =1\Rightarrow 4\mathrm{xy}={(\mathrm{x}+\mathrm{y})}^2$$$$\Rightarrow {\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2={(\mathrm{x}-\mathrm{y})}^2=0\Rightarrow \mathrm{x}=\mathrm{y}$$

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