Ps: I have reached the number of question I can propose so let this here please .And it's a work on open questions .
Does we have in radians :
$$\prod_{n=1}^{\infty}\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|=0$$
Some arguments in the right sense :
Let $n$ sufficiently large such that $0<\tan(n)<\sqrt{n}$ then it seems we have :
$$\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|<\frac{1}{\sqrt{n}} $$
Let $C$ be the number of solutions of $\tan(n)>\sqrt{n}$ then it seems we have for $n$ sufficiently large :
$$C<K\ln^2(n)$$
Where $K$ is superior or equal to one .
Moreover and again for $n$ sufficiently large such that $\tan(n)>\sqrt{n}$ it seems we have :
$$\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|<1+\frac{\left(n^{\frac{11}{10}}+1\right)}{\frac{n}{4}}$$
We have for $k+1<N$ and $N$ large enought :
$$-\left(N-k\right)^{\frac{1}{\sqrt{3}}}+\sum_{n=k}^{N}\frac{1}{\sqrt{n}}<0$$
Combining all this informations and using the theorem see https://math.stackexchange.com/questions/4536705/do-we-have-prod-n-1-infty-left1-frac-tan-leftn-right1n-left-tan my answer with telescoping (in the start)
It seems $\exists k,M$ two positives integers and $M$ sufficiently large :
$$\sum_{n=k}^{M}\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|<\left(M-k\right)\left|\left(1+\frac{1}{\left(M-k\right)^{1-\frac{1}{\sqrt{3}}}}+4\frac{\ln^{2}\left(M\right)-C}{(M-k)^{0.9}}\right)^{\frac{1}{M}}\left(1+\frac{k!\left(\tan\left(1\right)+1\right)}{M!\left(\tan\left(M+1\right)+1\right)}\right)\right|^{M}$$
Where $C$ is the number of solutions of $\tan(n)>\sqrt{n}$ for $n\leq M-k$
It leads currently for $k$sufficiently large and $k+1<M$ :
$$\prod_{n=k}^{M}\left|1+\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|\leq\left(1+\frac{1}{\left(M-k\right)^{1-\frac{1}{\sqrt{3}}}}+4\frac{\ln^{2}\left(M\right)-C}{(M-k)^{0.9}}\right)^{M-k}$$
Wich is very crude .
Side notes :
We can highly improving it replacing in the second inequality $<1+1/n$ or $1-1/n$ when we can
@FedorPetrov I stop to do maths here so be comprehensive the next question I can propose is in 3 month or more .Thanks you .
Also I'm thankfull to @Andreas for the bound $\tan(n)<n^{1.1}$ and @Ivan Kaznacheyeu for the this useful remark see https://math.stackexchange.com/questions/4542968/conjecture-about-n-tan-leftn1-right-tan-leftn-right1n-n-tann1n and https://math.stackexchange.com/questions/3499533/show-tann-nq-conjectured-q-1-1?noredirect=1&lq=1
Be gentle please .
As in we know recently that for $n$ an integer :
$$\tan(n)<n^{6.11}$$ then for $m$ large enought using the Cauchy criterion for convergence of infinite product :
$$|1+\prod_{n=m}^{m+k}\left|\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|-1|<\varepsilon$$
Then :
$$\prod_{n=m}^{\infty}\left|\frac{\tan\left(n\right)+1}{n\tan\left(n+1\right)+n}\right|=0$$
So now it's related to the symetric polynomials the problem being at least for $a_0$ to $a_6$
recalling the value above ($6.11$)
So now I'm thinking it diverges ...
