(b) $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}$

$\tan \theta = \frac{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{1 – 2{{\sin }^2}\frac{\theta }{2}}} = \frac{{2\tan \frac{\theta }{2}}}{{1 – {{\tan }^2}\frac{\theta }{2}}}$

$\left[ \begin{array}{l}{\rm{Using\,}}\,\,{\rm{sin}}\frac{\theta }{{\rm{2}}} = \sqrt {\frac{{x – 1}}{{2x}}} \\\therefore \,\,\cos \frac{\theta }{2} = \sqrt {1 – {{\sin }^2}\frac{\theta }{2}} = \sqrt {\frac{{x + 1}}{{2x}}} {\rm{\, \,and\,\,}}\,{\rm{tan}}\frac{\theta }{{\rm{2}}} = \frac{{\sqrt {x – 1} }}{{\sqrt {x + 1} }}\end{array} \right]$

$\therefore \,\,\,\tan \theta = \sqrt {{x^2} – 1} $.