Find minimum value of $frac{sec^4 alpha}{tan^2 beta}+frac{sec^4 beta}{tan^2 alpha}$

Question

I know this question has already answered here Then minimum value of $frac{sec^4 alpha}{tan^2 beta}+frac{sec^4 beta}{tan^2 alpha}$ but my answer is coming different.
i applied AM-GM directly to two fractions and by changing terms of sec and tan into sin and cos and simplifying a little we get that
$$frac{sec^4 alpha}{tan^2 beta}+frac{sec^4 beta}{tan^2 alpha}	geqfrac2{cosalpha cosbeta sinalpha sinbeta}geq2$$
but minimum value is coming 8 ???

Dhanvi Sreenivasan · Answer

$$frac{2}{cosalphacosbetasinalphasinbeta} = frac{8}{sin2alphasin2beta} geq 8$$
Now why did the previous inequality only give 2 whereas when we use this we get 8? Basically, $alpha$ and $beta$ are independent of each other, hence we can minimise the second expression by putting $alpha = beta = frac{pi}{4}$
On the other hand, in the expression you reduced to, both $sinalpha$ and $cosalpha$ cannot simultaneously be $1$, hence the product will actually have a different maximum value (which is $frac{1}{2}$)

Find minimum value of $frac{sec^4 alpha}{tan^2 beta}+frac{sec^4 beta}{tan^2 alpha}$

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