I used a Taylor series expansion of the product of trigonometric functions
Assume $$\arccos(x)\arcsin(x)=a_0+a_1x+a_2x^2+\ldots$$ The expansion is:
$$\arccos(x)\arcsin(x)=0+\frac{\pi}{2}x-x^2+\frac{\pi}{(2\cdot3!)}x^3-\frac{8}{4!}x^4-\frac{9\pi}{2\cdot5!}x^5-\frac{128}{6!}x^6+\dotsb$$
Substitute $x=1$, LHS becomes $0$:
$$0=\frac{\pi}{2}-1+\frac{\pi}{(2\cdot3!)}-\frac{8}{4!}-\frac{9\pi}{(2\cdot5)!}-\frac{128}{6!}+\dotsb$$
So, $$2\left(\frac{1}{2!}+\frac{(1\cdot2)^2}{4!}+\frac{(1\cdot2\cdot4)^2}{6!}+\dotsb\right)=\frac{\pi}{2}\left(\frac{(1\cdot-1)^2}{1!}+\frac{(1\cdot1)^2}{3!}+\frac{(1\cdot3)^2}{5!}+\frac{(1\cdot3\cdot5)^2}{7!}+\dotsb\right)$$
Rearranging gives $$\dfrac{\pi}{4}=\frac{\dfrac{(1)^2}{2!}+\dfrac{(1\cdot2)^2}{4!}+\dfrac{(1\cdot2\cdot4)^2}{6!}+\dfrac{(1\cdot2\cdot4\cdot6)^2}{8!}+\dotsb}{\dfrac{(-1)^2}{1!}+\dfrac{(-1\cdot1)^2}{3!}+\dfrac{(-1\cdot1\cdot3)^2}{5!}+\dfrac{(-1\cdot1\cdot3\cdot5)^2}{7!}+\dotsb}$$
I appreciate any alternate derivations for this result
Consider $$ \frac{\dfrac{(1)^2}{2!}+\dfrac{(1\cdot2)^2}{4!}+\dfrac{(1\cdot2\cdot4)^2}{6!}+\dfrac{(1\cdot2\cdot4\cdot6)^2}{8!}+\dotsb}{\dfrac{(-1)^2}{1!}+\dfrac{(-1\cdot1)^2}{3!}+\dfrac{(-1\cdot1\cdot3)^2}{5!}+\dfrac{(-1\cdot1\cdot3\cdot5)^2}{7!}+\dotsb} . $$ The numerator is $N(1)$, where $$ N(x) = \sum_{n=0}^\infty \frac{((2n)!!)^2}{(2n+2)!} x^{2n+2} = \frac{(\arcsin x)^2}{2} $$ The denominator is $D(1)$, where $$ D(x) = x + \sum_{n=1}^\infty \frac{((2n-1)!!)^2}{(2n+1)!} x^{2n+1} =\arcsin x . $$ So $$ N(1) = \frac{(\arcsin 1)^2}{2} = \frac{\pi^2}{8} \\ D(1) = \arcsin 1 = \frac{\pi}{2} $$ and our result is $$ \frac{N(1)}{D(1)} = \frac{\pi}{4} $$