Trace over $\mathbf{F}_2$ of $1/(\alpha + \alpha^{-1})$ where $\alpha^{2^n+1} = 1$ Let $n$ be an integer $\ge 2$.
Put $K = \mathbf{F}_{2^n}$ and $L = \mathbf{F}_{2^{2n}}$.
Let $\alpha$ be an element of $L$ such that $\alpha^{2^n+1} = 1$ and $\alpha \ne 1$. Using Sage, I have noticed that
$$ \mathrm{Tr}_{K/\mathbf{F}_2} \left( \frac{1}{\alpha + \alpha^{-1}} \right) = 1. $$
Is it correct ? If it is case, how to prove it ?
 A: Yes. This is correct.
Recall the following condition for solvability of a quadratic over a finite field of characteristic two:

The equation $x^2+ax+b=0$, $a,b\in K$, $a\neq0$ has two solutions in $K$, if and only if $Tr_{K/\Bbb{F}_2}(b/a^2)=0$.

Clearly
$$
x^2+ax+b=0\Longleftrightarrow \left(\frac x a\right)^2+\left(\frac x a\right)+\frac b{a^2}=0.
$$
So the highlighted result follows from the standard fact that $x^2+x+c=0$ has its solutions in $K$, iff $Tr_{K/\Bbb{F}_2}(c)=0$. 
Let $\omega=\alpha+\alpha^{-1}=\alpha+\alpha^{2^n}=Tr_{K/L}(\alpha).$ This is thus an element of $K$ (undoubtedly you knew this). It appears as a coefficient in the minimal polynomial of $\alpha$ over $K$, because
$$
(x-\alpha)(x-\alpha^{2^n})=x^2-(\alpha+\alpha^{2^n})x+\alpha^{2^n+1}=x^2+\omega x+1.
$$
This polynomial is a quadratic with coefficients in $K$. We also see that its zeros, $\alpha,1/\alpha$, are in $L\setminus K$. For if $\alpha\in K$, then we have both $\alpha^{2^n-1}=1$ and $\alpha^{2^n+1}=1$. Consequently also $\alpha^2=1$ contradicting the assumption $\alpha\neq1$.
Applying the above result to this polynomial says that
$$
Tr_{K/\Bbb{F}_2}(1/\omega^2)=1.
$$
But for all $z\in K$ we have $Tr_{K/\Bbb{F}_2}(z)=Tr_{K/\Bbb{F}_2}(z^2)$. The claim follows.
