Using the expression of $t^{\prime}$ as a function of $t$ and the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$, deduce
$$\forall n \in \mathbb{N}^{\star}, \quad \alpha_{2n+1} = \sum_{k=1}^{n} \binom{2n}{2k-1} \alpha_{2k-1} \alpha_{2n-2k+1}.$$