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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 think something is really wrong here. Take your first expression. We can look at it as a definition for $y$ as a function of $x$. Let $x=0$, then we have the following: $$2\sec^40-3\sec^20+2=\frac{\cos^20}{\cos^2y}.$$ 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 2 '13 at 0:04
Did you copy the problem down correctly? – Cameron Williams Jul 2 '13 at 0:06
@CameronWilliams I believe that the problem is correct. The issue that you're facing is the indeterminate form of $\frac{0}{0} $. (see my answer) – Calvin Lin Jul 2 '13 at 0:11
@CalvinLin well that's really unfortunate since there's nothing preventing $x=0,\pi$ in the original expression. – Cameron Williams Jul 2 '13 at 0:25
@CameronWilliams I agree. The question would be better phrased with $x\neq n\pi$, which would be all the stranger. – Calvin Lin Jul 2 '13 at 0:27
up vote 3 down vote accepted

(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.

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That relationship was quite impressive. I never would have guessed. – Cameron Williams Jul 2 '13 at 0:26
I can get the answer easily from here, but I can't see how to derivate this from the original question – chubakueno Jul 2 '13 at 0:29
@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 2 '13 at 0:29
@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 2 '13 at 0:31
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 2 '13 at 0:47

Some ideas:

$$1+\tan^2x=\sec^2x\implies (1+\tan^2x)^2=\sec^4x$$






Try now to end the exercise.

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I 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 2 '13 at 0:15
Yes, I ran into exactly that equation a while ago – chubakueno Jul 2 '13 at 0:16

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