If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:
$ (1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$