$\abs{\Gamma(1/2+it)}^2=\frac{2\pi}{e^{\pi t}+e^{-\pi t}}$

Put \[ s=\frac12+it. \] Then \(1-s=\frac12-it=\overline s\). Since \(\Gamma(\overline s)=\overline{\Gamma(s)}\), \[ \Gamma(s)\Gamma(1-s)=|\Gamma(s)|^2. \] The reflection formula gives \[ |\Gamma(1/2+it)|^2 =\frac{\pi}{\sin\pi(1/2+it)}. \] Now \[ \sin\left(\frac\pi2+i\pi t\right)=\cosh(\pi t) =\frac{e^{\pi t}+e^{-\pi t}}2. \] Therefore \[ |\Gamma(1/2+it)|^2 =\frac{2\pi}{e^{\pi t}+e^{-\pi t}}, \] and taking the positive square root yields \[ \boxed{|\Gamma(1/2+it)| =\sqrt{\frac{2\pi}{e^{\pi t}+e^{-\pi t}}}}. \]