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I wish to find an equation involving only $M$ and $N$ such that $x$ has been eliminated from these two equations.

$$M=\tan x+\cot x$$

$$N=\sec x-\cos x$$

I got this from my friend about a week ago and I have been trying this for a long time but still haven't been able to eliminate $x$.

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closed as unclear what you're asking by Surb, Claude Leibovici, Crostul, user223391, user7530 Sep 17 '16 at 21:06

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    $\begingroup$ are this two equations? $\endgroup$ – Dr. Sonnhard Graubner Sep 17 '16 at 9:28
  • $\begingroup$ Yes two equation m and n $\endgroup$ – Marvel Maharrnab Sep 17 '16 at 9:29
  • $\begingroup$ but it is not a system? $\endgroup$ – Dr. Sonnhard Graubner Sep 17 '16 at 9:31
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    $\begingroup$ $M =a/b +b/a$ and $N/b = 1/b^2 -1$, where $a = sin(x)$ and $b = cos(x)$. $\endgroup$ – Alex Silva Sep 17 '16 at 9:41
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    $\begingroup$ I think this question is reasonable. See my answer below. $\endgroup$ – mathworker21 Sep 17 '16 at 9:47
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First note that

$M^2 = \tan^2(x)+\cot^2(x)+2$

$N^2 = \sec^2(x)+\cos^2(x)-2$

So $M^2-N^2 = \cot^2(x)-\cos^2(x)+3$, which yields $M^2-N^2-3 = \frac{\cos^4(x)}{\sin^2(x)}$ (*).

Also note $\cot(x)M = \csc^2(x)$ and $\cos(x)N = \sin^2(x)$. Multiplying these two equations and squaring gives

$M^2N^2 = \frac{\sin^2(x)}{\cos^4(x)}$ (**).

Multiplying (*) and (**) gives $M^2N^2(M^2-N^2-3) = 1$.

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  • $\begingroup$ This was what i was talking about $\endgroup$ – Marvel Maharrnab Sep 17 '16 at 9:44

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