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$$ \frac{3}{1-\sin^6x-\cos^6x}=(\tan x + \cot x)^2$$ Need help with an identity I got for my high school homework. Can't seem to find a way to prove it. Please help with easiest way to do it. Thanks!

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    $\begingroup$ Unfortunately, I feel I had to flag this. This isn't a site that will solve your homework for you, and some effort or examples of your own thoughts must be shown. That way, we can help you—and others—the best we can. For more information, please read how to ask a good question and How to ask a homework question?. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! $\endgroup$ – Simply Beautiful Art Sep 17 '17 at 22:39
  • $\begingroup$ Before we can help, what have you tried so far? $\endgroup$ – NasuSama Sep 17 '17 at 22:40
  • $\begingroup$ I always do my homework by myself. I don't use the site for people to do homework for me... Whats the point of that? What knowledge do I get from that? I only refer to this site when I am stumped and really have no idea how to solve a problem. $\endgroup$ – RiktasMath Sep 17 '17 at 22:43
  • $\begingroup$ Please read how to ask a good question and How to ask a homework question?, as previously mentioned. $\endgroup$ – Simply Beautiful Art Sep 17 '17 at 22:45
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    $\begingroup$ How can you factorise $s^6+c^6$ ? $\endgroup$ – Donald Splutterwit Sep 17 '17 at 22:49
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The RHS reads

$$\left(\frac sc+\frac cs\right)^2=\frac{1^2}{c^2s^2}$$ hints you to rework the denominator of the LHS.

Terms $s^6+c^6$ can appear from the development of

$$1^3=(s^2+c^2)^3=s^6+3s^4c^2+3s^2c^4+c^6=s^6+c^6+3s^2c^2.$$

The rest is easy.

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  • $\begingroup$ How did you get this ? $\endgroup$ – RiktasMath Sep 17 '17 at 23:04
  • $\begingroup$ @RiktasMath: reduce to the common denominator and simplify. $\endgroup$ – Yves Daoust Sep 17 '17 at 23:07
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    $\begingroup$ $(\frac{s}{c}+\frac{c}{s})^2 = (\frac{c^2+s^2}{sc})^2 = (\frac{1}{sc})^2 = \frac{1}{(sc)^2} = \frac{1}{s^2c^2}$ For the second part, $3s^4c^2+3s^2c^4 = 3s^2c^2(s^2+c^2) = 3s^2c^2(1)$ $\endgroup$ – woogie Sep 17 '17 at 23:27
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    $\begingroup$ He makes use of a very convenient theorem called the Binomial Theorem, which gives the expansion of $$(x+y)^n$$ for any positive integer $n$. In the denominator, when you replace $1$ with $s^6+c^6+3s^2c^2$ they get cancelled out with the $-s^6-c^6$ that were already there on the LHS. $\endgroup$ – woogie Sep 17 '17 at 23:35
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    $\begingroup$ $\sin^2 x + \cos^2 x=1$ is a trig identity itself. Since it is always true for any value of x, we're free to exchange it wherever we need to. $\endgroup$ – woogie Sep 17 '17 at 23:56

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