Let
$$A_n(x):=\sum_{\sigma \in S_n} x^{1+\text{des}(\sigma)}$$
where $\text{des}(\sigma)$ is the number of descents of $\sigma$. These are called the Eulerian polynomials. I need to express $A_n(-1)$ in terms of Euler numbers. This is an exercise in Stanley's combinatorics vol 1 (exercise 135 chapter 1). In the solution the author suggests using the known fact that $$\sum_{d\geq 0} A_d(x) \frac{t^d}{d!}=\frac{1-x}{1-xe^{(1-x)t}}$$ and evaluate this equality in $x=-1$. Then the author suggests to compare the obtained expression with the Taylor-Maclaurin expansion of the tangent function.
My attempt
Evaluating in $x=-1$ $$\sum_{d\geq 0} A_d(-1) \frac{t^d}{d!}=\frac{2}{1+e^{2t}}$$ So $$\sum_{d\geq 0} A_d(-1)\frac{t^d}{d!}+\left(\sum_{d\geq 0} A_d(-1)\frac{t^d}{d!}\right)\left(\sum_{j\geq 0} 2^j \frac{t^j}{j!} \right)=2$$ Equating the coefficients on both sides I get messy equations and I really don't know how to use the Taylor series of the tangent.
We consider according to OPs calculation and formula (1.54) from Enumerative Combinatorics by R. P. Stanley the generating functions \begin{align*} \color{blue}{A(x)}&\color{blue}{:=\sum_{n\geq 0}A_n(-1)\frac{x^n}{n!}}\tag{1}\\ \color{blue}{\tan x}&\color{blue}{=\sum_{n\geq 0}E_{2n+1}\frac{x^{2n+1}}{(2n+1)!}}\tag{2} \end{align*}
Case $n$ even:
We obtain from (1) \begin{align*} \color{blue}{\sum_{n\geq 0}}&\color{blue}{A_{2n}(-1)\frac{x^{2n}}{(2n)!}}\\ &=\frac{1}{2}\left(A(x)+A(-x)\right)\\ &=\frac{1}{2}\left(\frac{2}{1+e^{2x}}+\frac{2}{1+e^{-2x}}\right)\\ &=\frac{1}{1+e^{2x}}+\frac{e^{2x}}{1+e^{2x}}\\ &\,\,\color{blue}{=1} \end{align*} and the claim (4) follows.
Case $n$ odd:
At first we derive a representation of the generating function with odd $A_n(-1)$. We obtain \begin{align*} \color{blue}{\sum_{n\geq 0}}&\color{blue}{A_{2n+1}(-1)\frac{x^{2n+1}}{(2n+1)!}}\\ &=\frac{1}{2}\left(A(x)-A(-x)\right)\\ &=\frac{1}{2}\left(\frac{2}{1+e^{2x}}-\frac{2}{1+e^{-2x}}\right)\\ &=\frac{1}{1+e^{2x}}-\frac{e^{2x}}{1-e^{2x}}\\ &\,\,\color{blue}{=\frac{1-e^{2x}}{1+e^{2x}}}\tag{5} \end{align*}
Now we recall Euler's formula and get \begin{align*} \tan x&=\frac{\sin x}{\cos x}=\frac{2}{2i}\,\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}} =-i\frac{e^{2ix}-1}{e^{2ix}+1}\\ &=i\frac{1-e^{2ix}}{1+e^{2ix}}\tag{6} \end{align*}
Combining (5) and (6) we obtain \begin{align*} \color{blue}{\frac{1-e^{2x}}{1+e^{2x}}} &=-i\tan (-ix)\\ &=-i\sum_{n\geq 0}E_{2n+1}\frac{(-ix)^{2n+1}}{(2n+1)!}\tag{$\to\ $(2)}\\ &\,\,\color{blue}{=\sum_{n\geq 0}E_{2n+1}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}} \end{align*} and the claim (3) follows.