Prove $\sum_{n=1}^{\infty} \arctan\left(\frac{1}{F_n}\right) \arctan\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{8}$

Let $a_n=\arctan(1/F_n)$, $b_n=a_n^2$ for even $n$, and $b_n=a_{n-1}a_{n+1}$ for odd $n$ (here we assume $a_0=\pi/2$, so that $b_1=\pi^2/8$). Now I claim that $\color{blue}{a_n a_{n+1}=b_n-b_{n+1}}$. We have $$a_{n-1}-a_{n+1}=\arctan\frac{F_{n+1}-F_{n-1}}{F_{n-1}F_{n+1}+1}=\arctan\frac{F_n}{F_n^2+1+(-1)^n}$$ (with $n=1$ checked separately), hence $a_{n-1}-a_{n+1}=a_n$ for odd $n$, and \begin{align*} n\text{ is odd }&\implies b_n-b_{n+1}=a_{n-1}a_{n+1}-a_{n+1}^2=(a_{n-1}-a_{n+1})a_{n+1},\\n\text{ is even }&\implies b_n-b_{n+1}=a_n^2-a_n a_{n+2}=a_n(a_n-a_{n+2}), \end{align*} giving $a_n a_{n+1}$ in both cases. Thus $\sum_{n=1}^\infty a_n a_{n+1}=\sum_{n=1}^\infty(b_n-b_{n+1})=b_1$.

COMMENT: May be this helps:

$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$

$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$

Squaring both sides of relations and subtract we get:

$$4\tan^{-1}x \tan^{-1}y=\left(\tan^{-1}\frac{x+y}{1-xy}\right)^2-\left(\tan^{-1}\frac{x-y}{1+xy}\right)^2$$

$x=\frac 1{F_n}$, and, $y=\frac 1{F_{n+1}}$

Now we use some relations between Fibonacci numbers, golden ratio $c_n$ is:

$c_n=\frac {F_{n+1}}{F_n}=\frac {F_nF_{n+1}}{F_n^2}\rightarrow F_nF_{n+1}=c_nF_n^2$

$$x+y=\frac1{F_n}+\frac1{F_{n+1}}=\frac{F_{n+2}}{c_n F_n^2}\rightarrow \frac{x+y}{1-xy}=\frac{F_nF_{n+1}F_{n+2}}{c_nF_n^2(F_nF_{n+1}-1)}$$

Similarly:

$$x-y=\frac1{F_n}-\frac1{F_{n+1}}=\frac{F_{n-1}}{c_n F_n^2}\rightarrow \frac{x-y}{1+xy}=\frac{F_{n-1}F_nF_{n+1}}{c_nF_n^2(F_nF_{n+1}+1)}$$

Now puting $F_1, F_2, F_3 \cdots$ gives us a numerical sum which must result in $\frac{\pi^2}8$.