<s>[INST] $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
 [/INST] a$\sqrt{1-\sqrt{x}}=a$
$\sqrt{1+\sqrt{x}}=b$
$ab=\sqrt{1-x}$
Now your expression will look something like :
$\frac{b+\sqrt{1+ab}}{a
+\sqrt{1-ab}}$
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