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If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$?

the answer is given as $-1$ in my book, but how do you arrive at this conclusion?

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don't understand, do you mean if $\cot a + 1/\cot a = 1$ then what is $\cot^2 a + 1/\cot^2 a$? – gt6989b Aug 28 '12 at 16:26
I'm wondering if the editing that's been done here is consistent with the original poster's intended meaning? – Michael Hardy Aug 28 '12 at 16:31
Note that $x+\frac 1 x$ cannot be equal to $1$ if $x$ is real. We have to bring in complex numbers here. – Michael Hardy Aug 28 '12 at 16:32
....and the question could have been phrased as follows: Why is it that if $x + \frac 1 x = 1$ then $x^2 + \frac{1}{x^2} = -1$? No need to mention cotangents at that point in the problem. – Michael Hardy Aug 28 '12 at 17:39
up vote 2 down vote accepted

Taking $x=\cot a$, $x+\frac{1}{x}=1\implies x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\frac{1}{x}=1-2=-1$

Alternatively, $x+\frac{1}{x}=1\implies x^2-x+1=0$

$x^2-x+1=0\implies x^3+1=0$



$x^{3m+1}+(\frac{1}{x})^{3m+1}=(x^3)^m\cdot x+\frac{1}{(x^3)^m\cdot x}=(-1)^m(x+\frac{1}{x})=(-1)^m$

$x^{3m+2}+(\frac{1}{x})^{3m+2}=(x^3)^m\cdot x^2+\frac{1}{(x^3)^m\cdot x^2}=(-1)^m(x^2+\frac{1}{x^2})=(-1)^m(-\frac{1}{x}-x)$ as $x^3=-1\implies x^2=-\frac{1}{x}$ and $\frac{1}{x^2}=x$

So,$x^{3m+2}+(\frac{1}{x})^{3m+2}=(-1)^{m+1}\cdot (x+\frac{1}{x})=(-1)^{m+1}$

If we put $m=0$, $x^2+\frac{1}{x^2}=(-1)^1=-1$

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Im sorry but i dont understand that very well. Can you go into more detail? – Aayush Agrawal Aug 28 '12 at 16:27
Are you aware of $a^2+b^2=(a+b)^2-2ab?$ – lab bhattacharjee Aug 28 '12 at 16:30
Omg, sadly i didnt D: i assume that there are more identities like that aswell?Can you please link me to a place where i can find them? Thanks a tonne! – Aayush Agrawal Aug 28 '12 at 16:32
@aayush You don't even need to be aware of this formula. Just square $(x+\frac{1}{x})$. – N. S. Aug 28 '12 at 16:36

Hint: $$ x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2. $$

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Why is that? How is that? How is there a -2? – Aayush Agrawal Aug 28 '12 at 16:31
Binomial formula: $$(a+b)^2=a^2+2ab+b^2 \Rightarrow a^2+b^2=(a+b)^2-2ab.$$ – Jyrki Lahtonen Aug 28 '12 at 16:33

Hint What do you get if you square $\cot(a)+\frac{1}{\cot(a)}$?

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