# Prove $\frac{3}{1-\sin^6x-\cos^6x}=(\tan x + \cot x)^2$.

$$\frac{3}{1-\sin^6x-\cos^6x}=(\tan x + \cot x)^2$$ 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!

• Before we can help, what have you tried so far? – NasuSama Sep 17 '17 at 22:40
• 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. – RiktasMath Sep 17 '17 at 22:43
• Please read how to ask a good question and How to ask a homework question?, as previously mentioned. – Simply Beautiful Art Sep 17 '17 at 22:45
• How can you factorise $s^6+c^6$ ? – Donald Splutterwit Sep 17 '17 at 22:49

$$\left(\frac sc+\frac cs\right)^2=\frac{1^2}{c^2s^2}$$ 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.

• How did you get this ? – RiktasMath Sep 17 '17 at 23:04
• @RiktasMath: reduce to the common denominator and simplify. – Yves Daoust Sep 17 '17 at 23:07
• $(\frac{s}{c}+\frac{c}{s})^2 = (\frac{c^2+s^2}{sc})^2 = (\frac{1}{sc})^2 = \frac{1}{(sc)^2} = \frac{1}{s^2c^2}$ For the second part, $3s^4c^2+3s^2c^4 = 3s^2c^2(s^2+c^2) = 3s^2c^2(1)$ – woogie Sep 17 '17 at 23:27
• 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. – woogie Sep 17 '17 at 23:35
• $\sin^2 x + \cos^2 x=1$ 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. – woogie Sep 17 '17 at 23:56