Question-215679

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

Answered by MATHEMATICSAM last updated on 14/Jan/25

sec^2 θ = ((4xy)/((x + y)^2 )) ⇒ cos^2 θ = (((x + y)^2 )/(4xy)) (x − y)^2 ≥ 0 ⇒ (x + y)^2 − 4xy ≥ 0 ⇒ (x + y)^2 ≥ 4xy ⇒ (((x + y)^2 )/(4xy)) ≥ 1 and its equal to 1 when x = y We know cos^2 θ ≤ 1 So cos^2 θ will be (((x + y)^2 )/(4xy)) when x = y. So sec^2 θ = ((4xy)/((x + y)^2 )) is only possible when x = y.

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

Answered by A5T last updated on 14/Jan/25

(x+y)^2 ≥4xy⇒sec^2 θ=((4xy)/((x+y)^2 ))≤((4xy)/(4xy))=1 But sec^2 θ=(1/(cos^2 θ))≥1 ⇒sec^2 θ≥1 and sec^2 θ≤1 which is only possible when secθ=1⇒4xy=(x+y)^2 ⇒x^2 −2xy+y^2 =(x−y)^2 =0⇒x=y

$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \geqslant\mathrm{4xy}\Rightarrow\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{4xy}}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }\leqslant\frac{\mathrm{4xy}}{\mathrm{4xy}}=\mathrm{1} \\ $$$$\mathrm{But}\:\mathrm{sec}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\geqslant\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{sec}^{\mathrm{2}} \theta\geqslant\mathrm{1}\:\mathrm{and}\:\mathrm{sec}^{\mathrm{2}} \theta\leqslant\mathrm{1}\:\mathrm{which}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible} \\ $$$$\mathrm{when}\:\mathrm{sec}\theta=\mathrm{1}\Rightarrow\mathrm{4xy}=\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} =\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{y} \\ $$

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