<s>[INST] Transforming trigonometric identities 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.
 [/INST] 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.
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