<s>[INST] How do I simplify $\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$? $$\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$$
I've had a go at putting everything over a common denominator in the form of:
$$\frac{\frac{\sqrt{1-x}\sqrt{1+x}+1}{\sqrt{1+x}}}{\frac{\sqrt{1-x}+1}{\sqrt{1-x}}} = \frac{\sqrt{1-x}(\sqrt{1-x^2}+1)}{(\sqrt{1-x}+1)\sqrt{1+x}}$$
Or multiplying the complex fraction by the conjugate of its base, but I'm getting nowhere.
Extensive searching hasn't made me realize the answer either— it has me baffled.
 [/INST] From where you left off, to remove all radicals from the denominator, we could multiply the top and bottom by $\sqrt{1+x}(1-\sqrt{1-x})$ -- the former factor to remove the factor of $\sqrt{1+x}$, and the latter being the conjugate of $1+\sqrt{1-x}$. This would result with
\begin{align*}
\frac{\sqrt{1-x}(\sqrt{1-x^2}+1)\sqrt{1+x}(1-\sqrt{1-x})}{(1+x)(1-(1-x))} &= \frac{\sqrt{1-x^2}(\sqrt{1-x^2}+1)(1-\sqrt{1-x})}{x(1+x)}
\\
&= \frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)}
\end{align*}
You could also expand the numerator, but I don't really think it seems "simpler".
\begin{align*}
\frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)} &= \frac{1-x^2+\sqrt{1-x^2}-\sqrt{1-x}+x^2\sqrt{1-x}-(1-x)\sqrt{1+x}}{x(1+x)} \\
&= \frac{(x^2-1)\sqrt{1-x}+(x-1)\sqrt{1+x}+\sqrt{1-x}\sqrt{1+x}+1-x^2}{x(1+x)}
\end{align*}
There are many different ways to write this expression.
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