Let $f(z)= 1/\cosh \pi z$. I would like to show that there exists a constant $A>0$ such that $$|f(x+iy)| \le \frac{A}{1+x^2} \; \text{for all} \; x\in \mathbb{R} \; \text{and} \; |y|<a$$ for any $|a| < \frac{1}{2}$.
To show this, I have first written $\cosh \pi z = e^{\pi z} \frac{1+e^{-2\pi z}}{2}$, and so $\frac{1}{\cosh \pi z} = e^{-\pi z} \frac{2}{1+e^{-2\pi z}}$. I know that $e^{-\pi z}$ would satisfy the above inequality, but I cannot deal with the denominator. How could I show this part?
Let $x \ge 0$. Note that $|e^{-\pi z}|=e^{-\pi x} \le \frac{1}{1+x^2}$ as $(1+x^2)e^{-\pi x}$ is positive decreasing on $[0, \infty)$, so we need to prove only that $|1+e^{-2\pi iz}| \ge C(a) >0$ when $|\Im z| < a <\frac{1}{2}$ as then we can take $A=\frac{2}{C(a)}$.
But $\Re {e^{-2\pi z}}=e^{-2\pi x}\cos 2\pi y$, so if $|a| \le \frac{1}{4}$, $\Re {e^{-2\pi z}} \ge 0$, so $|1+e^{-2\pi iz}| \ge 1$ hence we can take $C(a)=1$. If $\frac{1}{4} <a < \frac{1}{2}$, $\cos 2\pi y \ge \cos 2\pi a=\delta(a) >-1$ since for $|y| \le \frac{1}{4}, \cos 2\pi y \ge 0$, while if $\frac{1}{4} < |y| <a <\frac{1}{2}$, $\cos$ is negative and decreasing for the positive $y$ and obviously $\cos$ is even so the negative $y$ satisfies same inequality. Since $e^{-2\pi x} \le 1$, we get $\Re {e^{-2\pi z}} \ge \delta(a) >-1$, hence $|1+e^{-2\pi iz}| \ge 1+\delta(a) >0$, so we take $C(a)=1+\delta(a)$.
(note that if $0 \le c \le 1, w<0, cw \ge w$, so $e^{-2\pi x}\cos 2\pi y \ge e^{-2\pi x}\cos 2\pi a \ge \cos 2\pi a$, when $\cos 2\pi a <0$)
For $x$ negative one can use that the cosine hyperbolic is even so the inequality is same as for $-x$ (or take $e^{-\pi z}$ factor and repeat)