If $y>0$, prove that $|\cot(x+iy)+i| < \frac{e^{-2y}}{1-e^{-2y}}$

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If $y>0$, prove that $$|\cot(x+iy)+i| < \frac{e^{-2y}}{1-e^{-2y}}$$

I'm new on this. By now I got this

$$\cot(z)=\frac{\cos(z)}{\sin(z)}=\frac{e^{iz}+e^{-iz}}{2}.\frac{2i}{e^{iz}-e^{-iz}} = \frac{(e^{iz}+e^{-iz})i}{e^{iz}-e^{-iz}}$$

From here I got no idea what to do.

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Use that $$\cot(x+iy)={\frac {-1-i{\rm coth} \left(y\right)\cot \left( x \right) }{-i {\rm coth} \left(y\right)+\cot \left( x \right) }} $$