I'm working on Exercise 9, Section 19 of Morandi's Field and Galois Theory:
If $K=\mathbb{R}(x,\sqrt{1+x^2})$, show that there is a $t\in K$ with $K=\mathbb{R}(t)$.
Trying to deduce a workable $t$, I assume if $K=\mathbb{R}(t)$ for some $t$, then I can write $x=g(t)=u(t)/a(t)$ and $\sqrt{1+x^2}=f(t)=v(t)/b(t)$ for some rational functions $g$ and $f$ in lowest terms for some polynomials $u,v,a,b$. This gives a relation $$ f(t)^2=1+x^2=1+g(t)^2 $$ and then $$ a^2v^2=b^2(a^2+u^2). $$
Since $v$ and $b$ are coprime, $b^2\mid a^2$, and similarly $a^2\mid b^2$, so $a$ and $b$ are associate in $\mathbb{R}[t]$. Up to scaling $u$ and $v$ a unit, I can assume $a=b$ to put $f$ and $g$ over a common denominator. This leads to an equation $$ v(t)^2=a(t)^2+u(t)^2 $$ but I'm having a hard time coming up with an appropriate $t$ in terms of $x$ to show $K=\mathbb{R}(t)$. I thought maybe the exercise has a typo, since the book is known to have many errata, and tried disproving it with a method like that found here, but I don't think the same approach works since there isn't a related elliptic curve?
I am wondering if maybe the exercise means to show $\mathbb{R}(x,\sqrt{1-x^2})=\mathbb{R}(t)$ for some $t$? Then for instance $t=\frac{\sqrt{1-x^2}}{1-x}$ gives $$ t^2=\frac{1-x^2}{(1-x)^2}=\frac{1+x}{1-x} $$ which is an invertible linear fractional transformation, so $x\in\mathbb{R}(t^2)\subseteq\mathbb{R}(t)$ and ultimately gives $\mathbb{R}(x,\sqrt{1-x^2})=\mathbb{R}(t)$. I was suspecting this since $x$ and $y:=\sqrt{1-x^2}$ parametrize the circle $x^2+y^2=1$ which seems more standard when considering geometric examples, instead of a hyperbola.
