This is exercise 7 of Section 19 on Transcendence Degrees in Morandi's Field and Galois Theory.
Let $K=\mathbb{C}(x)(\sqrt{-1-x^2})$. Show that $[K:\mathbb{C}(x)]=2$, and show that $K=\mathbb{C}(t)$ if $t=(i-x)^{-1}\sqrt{(-1-x^2)}/(i-x)$.
It was easy for me to show $[K:\mathbb{C}(x)]=2$, but I can't reach the conclusion $K=\mathbb{C}(t)$. The formula for $t$ above is exactly as it is typeset in the book, but I wonder if it really means $$ t=(i-x)^{-1}\sqrt{\frac{-1-x^2}{i-x}},\qquad (*) $$ otherwise why wouldn't it be written as $t=\sqrt{(-1-x^2)}/(i-x)^2$ or even $t=(i-x)^{-2}\sqrt{(-1-x^2)}$?
Assuming the form in $(*)$, I was able to determine $t\notin\mathbb{C}(x)$, since if $t=\frac{f}{g}$ for some $f,g\in\mathbb{C}[x]$, squaring and rearranging gives $g^2\cdot (-1-x^2)=f^2\cdot (i-x)^3$, a contradiction by looking at degrees. I get a tower $$ \mathbb{C}(x)\subsetneq\mathbb{C}(x,t)\subseteq K $$ so $\mathbb{C}(x,t)=K$. This just brings me back to wanting to show $x\in\mathbb{C}(t)$. But this seems obvious from the get go, so I don't know how to proceed, and I haven't used anything about transcendence degrees.