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Question Number 177793 by cortano1 last updated on 09/Oct/22

Determine the minimum value ((sec^4 α)/(tan^2 β)) + ((sec^4 β)/(tan^2 α)) over all α,β ≠ ((kπ)/2) , k ε Z

$$\:\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value} \\ $$$$\:\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\: \\ $$$$\:\:\mathrm{over}\:\mathrm{all}\:\alpha,\beta\:\neq\:\frac{\mathrm{k}\pi}{\mathrm{2}}\:,\:\mathrm{k}\:\varepsilon\:\mathbb{Z}\: \\ $$

Answered by mahdipoor last updated on 09/Oct/22

get tan^2 x=z⇒ sec^4 x=((1/(cos^2 x)))^2 =(1+tan^2 x)^2 =(1+z)^2 ⇒⇒ ⇒tan^2 β=m⇒sec^4 β=(1+m)^2 ⇒tan^2 α=n⇒sec^4 α=(1+n)^2 ⇒⇒ f(α,β)=f(m,n)=(((1+m)^2 )/n)+(((1+n)^2 )/m) m,n≠0 { (((∂f/∂m)=((2(1+m))/n)−(((1+n)^2 )/m^2 )=0)),(((∂f/∂n)=−(((1+m)^2 )/n^2 )+((2(1+n))/m)=0)) :} ⇒m=n=1 min f=f(1,1)=8

$${get}\:\:{tan}^{\mathrm{2}} {x}={z}\Rightarrow \\ $$$${sec}^{\mathrm{4}} {x}=\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}\right)^{\mathrm{2}} =\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} =\left(\mathrm{1}+{z}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow \\ $$$$\Rightarrow{tan}^{\mathrm{2}} \beta={m}\Rightarrow{sec}^{\mathrm{4}} \beta=\left(\mathrm{1}+{m}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{tan}^{\mathrm{2}} \alpha={n}\Rightarrow{sec}^{\mathrm{4}} \alpha=\left(\mathrm{1}+{n}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow \\ $$$${f}\left(\alpha,\beta\right)={f}\left({m},{n}\right)=\frac{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} }{{n}}+\frac{\left(\mathrm{1}+{n}\right)^{\mathrm{2}} }{{m}}\:\:\:{m},{n}\neq\mathrm{0} \\ $$$$\begin{cases}{\frac{\partial{f}}{\partial{m}}=\frac{\mathrm{2}\left(\mathrm{1}+{m}\right)}{{n}}−\frac{\left(\mathrm{1}+{n}\right)^{\mathrm{2}} }{{m}^{\mathrm{2}} }=\mathrm{0}}\\{\frac{\partial{f}}{\partial{n}}=−\frac{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} }{{n}^{\mathrm{2}} }+\frac{\mathrm{2}\left(\mathrm{1}+{n}\right)}{{m}}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow{m}={n}=\mathrm{1} \\ $$$${min}\:{f}={f}\left(\mathrm{1},\mathrm{1}\right)=\mathrm{8} \\ $$

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