Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:

$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$

(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?

The hint
$$
\frac{ \sqrt{1+\sin(x)}+\sqrt{1+\cos(x)} }{ \sqrt{1-\sin(x)}+\sqrt{1-\cos(x)} }=1+\sqrt2 \quad \text{for } x\in [0,\pi/2]
$$
is spot on.
We have
$$
\begin{align}
\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}}
&=
\sqrt{10}\left(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}}\right)
\\&=
\sqrt{10}\left(\sqrt{1+\sin(t)}+\sqrt{1+\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sin(t)}+\sqrt{1-\cos(t)}\right)
\\&=
\sqrt{10}(1+\sqrt 2)\left(\sqrt{1-\sqrt{0.01}}+\sqrt{1-\sqrt{0.99}}\right)
\\&=
(1+\sqrt 2)\left(\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{99}}\right)
\end{align}
$$
for some $t$.
Therefore, we can rearrange the numerator to be $1+\sqrt 2$ times the denominator. (You need to handle the middle terms separately but it works out the same.)
