An uncommon continued fraction of $\frac{\pi}{2}$

The result comes from a rather slight modification of Euler's continued fraction , $$a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+....=\cfrac{a_0}{1-\cfrac{a_1}{1+a_1-\cfrac{a_2}{1+a_2-\cfrac{a_3}{\cdots}}}}\label{1}\tag{1}$$

Note that for a product $a_0a_1a_2\cdots$; we can represent them as, $$\begin{align}a_{1}a_{2}a_{3}...&=a_{1}+a_{1}\left(a_{2}-1\right)+a_{1}a_{2}\left(a_{3}-1\right)+\cdots\label{2}\tag{2}\end{align}$$ With \eqref{2} in \eqref{1}, $$a_{1}a_{2}a_{3}a_{4}\cdots=1+\cfrac{a_{1}-1}{1-\cfrac{a_{1}\left(a_{2}-1\right)}{a_{2}a_{1}-1-\cfrac{\left(a_{1}-1\right)a_{2}\left(a_{3}-1\right)}{a_{3}a_{2}-1-\cfrac{\left(a_{2}-1\right)a_{3}\left(a_{4}-1\right)}{\cdots}}}}\label{3}\tag{3}$$

Now apply the Wallis product, $$\frac{\pi}{2}=\frac{2\cdot2}{1\cdot3}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{6\cdot6}{5\cdot7}\cdots$$ in \eqref{3} with $a_1=2/1$, $a_2=2/3$ ... to get the first result

The second result is not obtained via the Wallis product, but from the Leibniz series for $\pi/4$ using \eqref{1}

We know that $$\sin^{-1}x=\int\underbrace{\color{red}{\frac{1}{\sqrt{1-x^2}}}}_{\text{apply binomial theorem}}dx=\int1+\sum_{n=0}^{\infty}\frac{2n-1}{2^{n+1}}x^{2n}dx$$and if you solve this further you'll get $$\sin^{-1}x=x+ \frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+......$$ $$= x+\frac{1}{2}\cdot\frac{x^3}{3}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{x^5}{5}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{x^7}{7}+......$$ $$=x+x\left(\frac{x^2}{2\cdot3}\right)+x\left(\frac{x^2}{2\cdot3}\right)\left(\frac{(3x)^2}{4\cdot5}\right)+.........$$ OR $$\sin^{-1}x=x\left(1+\sum_{n=0}^{\infty}\prod_{i=0}^{n}\frac{(2i+1)^2x^2}{(2i+2)\cdot(2i+3)}\right)$$

Now according to Euler's formula for Continued Fraction , i.e. $$S=a\left(1+\sum_{i=1}^{\infty}\prod_{j=1}^{i}r_j\right)=\large\frac{a}{1-\frac{r_1}{1+r_1-\frac{r_2}{1+r_2-\frac{r_3}{......}}}}$$

So , $$\sin^{-1}x=x\left(1+\sum^{\infty}_{n=1}\prod_{i=1}^{n}\frac{(2i-1)^2x^2}{(2i)\cdot(2i+1)}\right)$$ $$=x\Large\left(\frac{1}{1-\frac{\frac{x^2}{2\cdot3}}{1+\frac{x^2}{2\cdot3}-\frac{\frac{3^2x^2}{4\cdot5}}{1+\frac{3^2x^2}{4\cdot5}-.......}}}\right)$$

Since , $$\sin^{-1}1=\frac{\pi}{2}$$ Now just put $x=1$ in the above continued fraction to get to your answer .