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I am trying to show that for $z=x+iy$ where $|y|\geq 1$, $$|\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}.$$

Here is my work so far: \begin{align*} |\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\ &=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\ &=\left\vert\frac{\exp(-2\pi y)+\exp(-2i\pi x)}{\exp(-2\pi y)-\exp(-2i\pi x)}\right\vert\\ &\leq\frac{1+\exp(-2\pi y)}{1-\exp(-2\pi y)}, \end{align*} where the last step is from the triangle inequality in the numerator and the reverse triangle inequality in the denominator. It appears that I have lost the absolute value sign of $y$; did I do anything wrong?

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The last step in your calculation is wrong for negative $y$. You have $$ |\exp(-2\pi y) + \exp(-2i\pi x)| \le 1 + \exp(-2\pi y) \\ |\exp(-2\pi y) - \exp(-2i\pi x)| \ge 1 - \exp(-2\pi y) $$ but the right-hand side of the second inequality is negative if $y < 0$, so that you cannot divide the inequalities.

But your calculation is correct for $y > 0$. For negative $y$ you can write \begin{align*} |\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\ &=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\ &=\left\vert\frac{\exp(2i\pi x)+\exp(2\pi y)}{\exp(2i\pi x)-\exp(2\pi y)}\right\vert\\ &\leq\frac{1+\exp(2\pi y)}{1-\exp(2\pi y)} = \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)} \end{align*} behause then $1-\exp(2\pi y) > 0$.

Alternatively, use that $|\cot |$ is an even function, so that for $y < 0$ $$ |\cot \pi (x+iy)| = |\cot \pi (-x-iy)| \leq\frac{1+\exp(-2\pi (-y))}{1-\exp(-2\pi (-y))} = \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)} \, . $$

Note that this proves $$ |\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}. $$ for all $z=x+iy$ with $y \ne 0$, not only for $|y| \ge 1$.

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