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 $x$.


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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $M =a/b +b/a$ and $N/b = 1/b^2 -1$, where $a = sin(x)$ and $b = cos(x)$. $\endgroup$ – Alex Silva Sep 17 '16 at 9:41
  • 3
    $\begingroup$ I think this question is reasonable. See my answer below. $\endgroup$ – mathworker21 Sep 17 '16 at 9:47

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 $M^2N^2(M^2-N^2-3) = 1$.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.