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

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 $$\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha} \geq\frac2{\cos\alpha \cos\beta \sin\alpha \sin\beta}\geq2$$

but minimum value is coming 8 ???

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    $\begingroup$ Your bound $2/(\cos\alpha\sin\alpha\cos\beta\sin\beta)\geq 2$ is too weak. You can tighten that to $\geq 8$ by using double angle formula for sine. $\endgroup$ Jul 23 '20 at 3:29
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    $\begingroup$ Right, so the expression is always $\ge2$. But $2$ isn't the minimum value, since there are no $\alpha$ and $\beta$ making the expression $2$ (since actually it's always $\ge8$). $\endgroup$ Jul 23 '20 at 3:29
  • $\begingroup$ @AnginaSeng but how we know that there are no $α$ and $β$ making the expression $2$ , and how we get to know that 8 is correct minimum in the same sense ? $\endgroup$
    – Ishan
    Jul 23 '20 at 3:31
  • $\begingroup$ @Ishan see math.stackexchange.com/questions/1966853/… $\endgroup$ Jul 23 '20 at 3:31
  • $\begingroup$ @AnginaSeng thanks,but i have already seen it and also i have mention same link in my question,but i am asking how we know that 2 is not minimum ? $\endgroup$
    – Ishan
    Jul 23 '20 at 3:32
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$$\frac{2}{\cos\alpha\cos\beta\sin\alpha\sin\beta} = \frac{8}{\sin2\alpha\sin2\beta} \geq 8$$

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 $\sin\alpha$ and $\cos\alpha$ cannot simultaneously be $1$, hence the product will actually have a different maximum value (which is $\frac{1}{2}$)

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  • $\begingroup$ sorry,but still did not get it , first of all if i have came to minimum 2 why i think to come on 8 ? and also i did not use fact that both sin and cos are simultaneously 1 ,i just use that they are always <1 whatever a and b is. $\endgroup$
    – Ishan
    Jul 23 '20 at 3:40
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    $\begingroup$ @Ishan You cannot maximize the expression $\cos x \cos y \sin x \sin y$ by simply saying that the maximum value of $\sin$ is always $1$ and the maximum value of $\cos$ is also always $1$ so the maximum value of their product will be $=1$. This is wrong because while you're figuring out the angle to maximize $\sin$, the cosine part of your expression is simultaneously getting minimized. If you want to think about the maximum of $\sin \alpha \cos \beta \sin \gamma \cos \delta$, it would definitely be $=1$ as the variables are independent, but in your case you have some common variables. $\endgroup$
    – Nikunj
    Jul 23 '20 at 3:48
  • $\begingroup$ @Nikunj got it, thankyou very much $\endgroup$
    – Ishan
    Jul 23 '20 at 3:59
  • $\begingroup$ @Nikunj i was doing another problem and its solution here artofproblemsolving.com/community/c6h212557p1172899 is contradicting what you are saying!!! i request you to please see the first solution in the link $\endgroup$
    – Ishan
    Jul 26 '20 at 2:08
  • $\begingroup$ @Nikunj in the end he says that $ \sum {\cos (x - y)} > 3$ which is false and reason provided there is same that i am mentioning above but there also we have some common variables inside cos ?? i am confused .. $\endgroup$
    – Ishan
    Jul 26 '20 at 2:09

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