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


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

$Determine the minimum value$ $\frac{\sec^{4} α}{\tan^{2} β} + \frac{\sec^{4} β}{\tan^{2} α}$ $over all α, β \neq \frac{k π}{2}, k ε Z$
	Answered by mahdipoor last updated on 09/Oct/22
	$get tan^2 x=z⇒ sec^4 x=((1/(cos^2 x)))^2 =(1+tan^2 x)^2 =(1+z)^2 ⇒⇒ ⇒tan^2 β=m⇒sec^4 β=(1+m)^2 ⇒tan^2 α=n⇒sec^4 α=(1+n)^2 ⇒⇒ f(α,β)=f(m,n)=(((1+m)^2 )/n)+(((1+n)^2 )/m) m,n≠0 { (((∂f/∂m)=((2(1+m))/n)−(((1+n)^2 )/m^2 )=0)),(((∂f/∂n)=−(((1+m)^2 )/n^2 )+((2(1+n))/m)=0)) :} ⇒m=n=1 min f=f(1,1)=8$
	$g e t {t a n}^{2} x = z \Rightarrow$ ${s e c}^{4} x = {(\frac{1}{{c o s}^{2} x})}^{2} = {(1 + {t a n}^{2} x)}^{2} = {(1 + z)}^{2}$ $\Rightarrow\Rightarrow$ $\Rightarrow {t a n}^{2} β = m \Rightarrow {s e c}^{4} β = {(1 + m)}^{2}$ $\Rightarrow {t a n}^{2} α = n \Rightarrow {s e c}^{4} α = {(1 + n)}^{2}$ $\Rightarrow\Rightarrow$ $f (α, β) = f (m, n) = \frac{{(1 + m)}^{2}}{n} + \frac{{(1 + n)}^{2}}{m} m, n \neq 0$ ${\begin{cases} \frac{\partial f}{\partial m} = \frac{2 (1 + m)}{n} - \frac{{(1 + n)}^{2}}{m^{2}} = 0 \\ \frac{\partial f}{\partial n} = - \frac{{(1 + m)}^{2}}{n^{2}} + \frac{2 (1 + n)}{m} = 0 \end{cases}$ $\Rightarrow m = n = 1$ $m i n f = f (1, 1) = 8$
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