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In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?

$$frac{{displaystyle sum_{n=1}^{99}} sqrt{10+sqrt{n}}}{{displaystyle sum_{n=1}^{99}} sqrt{10-sqrt{n}}}$$

This can be done using an identity I learned from math.SE.

For any $a > 0$ and $0 le b le a^2$, we have
$$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$

To prove this identity, just take squares on both sides and use the fact

$$begin{align}
left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
&= 2left(a + sqrt{a^2-b}right)end{align}$$

Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us

$$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
= sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
= sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
This leads to
$$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
= frac{1}{sqrt{2}-1} = sqrt{2}+1
$$
Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.