(b) Given, ${\rm{cosec}}\theta = \frac{{p + q}}{{p – q}}$

==> $\frac{1}{{\sin \theta }} = \frac{{p + q}}{{p – q}}$

Apply componendo and dividendo

$\frac{{1 + \sin \theta }}{{1 – \sin \theta }} = \frac{{p + q + p – q}}{{p + q – p + q}}$

==> ${\left\{ {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} – \sin \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$

==> ${\left\{ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 – \tan \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$

==> ${\tan ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{p}{q}$

==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{q}{p}$

Note : $\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \sqrt {\frac{q}{p}} \,{\rm{only,}}\,\,{\rm{if}}$

$\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) > 0$.