If $\mu =\sec \theta +\tan \theta$then  $\tan \theta$in terms of  $\mu$is___  

Given

$\frac{\begin{array}{l}\sec \theta +\tan \theta =\mu \\ \sec \theta -\tan \theta =\frac{1}{\mu }\end{array}}{\begin{array}{l}2\tan \theta =\mu -\frac{1}{\mu }\\ 2\tan \theta =\frac{{\mu }^2-1}{\mu }\\ \tan \theta =\frac{{\mu }^2-1}{2\mu }\end{array}}$