Using the result that for all $x \in ]0,1[$:
$$\frac{\pi}{\sin(\pi x)} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+x} + \sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1-x},$$
deduce that, for $x \in ]-\frac{1}{2}, \frac{1}{2}[$:
$$\frac{\pi}{\cos(\pi x)} = \sum_{k=0}^{+\infty} \left(\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^{2k+1}}\right) 2^{2k+2} x^{2k}.$$