The equation $$\ln \left|\frac{\sin x}{\cos x - 1}\right| = \ln \left|\frac{\cos x + 1}{\sin x}\right|\tag{1}\;,$$ is equivalent to $$\left|\frac{\sin x}{\cos x - 1}\right| = \left|\frac{\cos x + 1}{\sin x}\right|\;,\tag{2}$$ which is equivalent to $$|\sin x|^2=|\cos x-1||\cos x+1|\;,\tag{3}$$ for all values of $x$ for which $(1)$ makes sense (i.e., $x\ne n\pi$ for integer $n$). Thus, $(1)$ is equivalent to $$\sin^2 x=|\cos^2x-1|\tag{4}$$ for all values of $x$ for which $(1)$ makes sense.
But $0\le\cos^2 x\le 1$ for all $x$, so $|\cos^2x-1|=1-\cos^2x$, and $(4)$ is equivalent to the familiar Pythagorean identity for all $x$. As noted, the steps are reversible for all $x$ for which $(1)$ makes sense, so $(1)$ is indeed an identity.