Consider the operator $\frac D{e^D-1}$ which we will call "shadow":
$$\frac {D}{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } e^{-iwx}\frac{-iw}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^{i t w} f(t) \, dt \, dw$$
The integrals here should be understood as Fourier transforms.
Now, intuitively, why the following?
$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1\pi\ln \left(\frac{x+1/2 +\frac{z}{\pi }}{x+1/2 -\frac{z}{\pi }}\right)\right]\right|_{x=0}=\tan z$$
There are other examples where shadow converts trigonometric functions into inverse trigonometric, logarithms to exponents, etc:
$$\left.\frac {D_x}{e^{D_x}-1} \left[\frac1{\pi }\ln \left(\frac{x+1-\frac{z}{\pi }}{x+\frac{z}{\pi }}\right)\right]\right|_{x=0}=\cot z$$
Best Answer
This is basically a lightly transformed version of Euler's cotangent partial fraction expansion $$ \pi \cot(\pi z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{1}{z-n} + \frac{1}{z+n}$$ (the log derivative of his famous sine product formula $\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \big(1-\frac{z^2}{n^2}\big)$). By telescoping series one can rewrite this as $$ \pi \cot(\pi z) = \sum_{n=0}^\infty \frac{1}{z-n-1} + \frac{1}{z+n}.$$ By Taylor's theorem, $e^{nD_x}$ is the operation of translation by $n$, so formally by geometric series we have $$ \left.\frac{1}{1-e^{D_x}} f\, \right|_{x=0} = \sum_{n=0}^\infty \left.e^{nD_x} f\right|_{x=0} = \sum_{n=0}^\infty f(n)$$ (which incidentally helps explain the Euler-Maclaurin formula) and so $$ \pi \cot(\pi z) = \left.\frac{1}{1-e^{D_x}} \left(\frac{1}{z-x-1} + \frac{1}{z+x}\right) \right|_{x=0}$$ or equivalently $$ \pi \cot(\pi z) = - \left.\frac{D_x}{1-e^{D_x}} \ln \frac{x+z}{x+1-z} \right|_{x=0}.$$ This gives your identities after some simple rearrangements (and replacing $z$ with either $z/\pi$ or $z/\pi + 1/2$).
The primary reason for Euler's partial fraction identity is that the poles and residues of the cotangent function are easily identified and computed. The reason they can be collapsed into an expression involving the summation operator $\frac{1}{1-e^{D_x}}$ is that these poles and residues enjoy a translation invariance, which ultimately comes from the periodicity of the cotangent function. I would imagine there are similar identities for the Weierstrass $\wp$ function, which is doubly periodic with very specific pole behavior.