The problem goes like this: If $$N=2\sec^4x-3\sec^2x+2=\frac{\cos^2x}{\cos^2y}$$ Calculate the equivalent of $$M=2\tan^4x+3\tan^2x+2$$ The alternaties I have are: $$\frac{\tan^2x}{\tan^2y},\mbox{ }\frac{\tan^2y}{\tan^2x},\mbox{ }\frac{\tan^2y}{\sec^2x},\mbox{ }\frac{\sec^2y}{\tan^2x},\mbox{ }\frac{\sec^2x}{\tan^2y}$$ The first thing I tried was to "build" the value of $N$ and then use it for $M$ $$\sec^2x=1+\tan^2x\\ \sec^4x=1+2\tan^2x+\tan^4x\\ 2\sec^4x=2+4\tan^2x+2\tan^4x\\ -3\sec^2x=-3-3\tan^2x\\ 2\sec^4x-3\sec^2x+2=1+\tan^2x+2\tan^4x\\ \frac{\cos^2x}{\cos^2y}=1+\tan^2x+2\tan^4x\\ \frac{\cos^2x}{\cos^2y}+2\tan^2x+1=2+3\tan^2x+2\tan^4x\\ \frac{\cos^2x}{\cos^2y}+2\tan^2x+1=M$$ But then I can't transform the final equation to one of the alternatives, even after trying a massive substition of $\cos^2y$ it didn't helped too much. Any hints or ideas are greatly appreciated.
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(I do not like this solution, and do not have a good explanation for the observation, which is based on the potential answers. There is no hint that I can give, which doesn't give away the entire game.)
Observe that $$M \sin^2x - N = - \cos^2x ,$$
which you can check in Wolfram (see alternate forms).
Hence, the answer is __ (fill in the blank yourself).
The issue that Cameron was having arose, because we have to divide by $\sin^2 x$ (which is 0 when $x=0$) in determining the answer.
Calvin Lin
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That relationship was quite impressive. I never would have guessed. – Cameron Williams Jul 02 '13 at 00:26
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I can get the answer easily from here, but I can't see how to derivate this from the original question – chubakueno Jul 02 '13 at 00:29
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@CameronWilliams Neither would I. I'm still trying to figure out various other such identities. I've seen some before, with very little explanation of what is happening apart from pure dumb luck of cancellation. This is one of those questions of the type "There isn't much to try, so let's try these few things, and hey presto it works". I can't explain the intuition behind it either, nor motivate how to arrive at similar equations. – Calvin Lin Jul 02 '13 at 00:29
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@chubakueno I freely admit that I have no idea how to obtain this (and hence the initial spew). What intrigued me was that the fourth answer choice, which says that $M= \frac{N}{\sin^2x}$, so I tried the LHS, and out popped the RHS. Had there been no answer choices, I would be stuck too. – Calvin Lin Jul 02 '13 at 00:31
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This was on the "challenge"["Tú puedes" in spanish] part of my book .As I was aware that having no objectives to "get" from the initial problem would be too broad, I decided to publish the alternatives. Normally the questions are much easier than this one, so I left this question for the final. It really surprises me that it was so difficult. – chubakueno Jul 02 '13 at 00:47
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Some ideas:
$$1+\tan^2x=\sec^2x\implies (1+\tan^2x)^2=\sec^4x$$
But
$$(1+\tan^2x)^2=1+2\tan^2x+\tan^4x$$
So
$$2\tan^4x+3\tan^2x+2=2(\tan^4x+2\tan^2x+1)-\tan^2x=2\sec^4x-\tan^2x=$$
$$=2\sec^4x-\sec^2x+1$$
Try now to end the exercise.
DonAntonio
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2I believe that OP has tried something similar to that. I'd be interested if you can complete it this way (I couldn't). – Calvin Lin Jul 02 '13 at 00:15
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$$2\sec^40-3\sec^20+2=\frac{\cos^20}{\cos^2y}.$$
Upon simplification, we see that $\cos^2y = 1$ at $x=0$. Thus $y = 0,\pi$ (of course it could be any multiple of $\pi$, but these are the only two values that really matter).
If we let $x=0$ in your expression for $M$, we have that $M = 2$, but in none of the possible answers can $M$ come out to be $2$.
– Cameron Williams Jul 02 '13 at 00:04