$\cos ec\,\theta  = \frac{{p + q}}{{p – q}},$ $\sin \theta  = \frac{{p – q}}{{p + q}}$

$\cos \theta$

$=\pm \sqrt{1-\sin ^{2} \theta}=\sqrt{1-\left(\frac{p-q}{p+q}\right)^{2}}=\frac{2 \sqrt{p q}}{(p+q)}$

$\left|\cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right|=\frac{\cot \frac{\pi}{4} \cot \frac{\theta}{2}-1}{\cot \frac{\pi}{4}+\cot \frac{\theta}{2}}$

$=\frac{\cot \frac{\theta}{2}-1}{\cot \frac{\theta}{2}+1}=\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}$

On rationalizing denominator, we get

${\left( {\frac{{\cos \frac{\theta }{2} – \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}} \right)}$  ${\left( {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}} \right)}$

$ = \,\frac{{\cos \,\theta }}{{{{\sin }^2}\frac{\theta }{2} + {{\cos }^2}\frac{\theta }{2} + 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}$

$ = \left| {\frac{{\cos \theta }}{{1 + \sin \theta }}} \right|\, = \,\frac{{2\sqrt {pq} /(p + q)}}{{1 + \,\frac{{(p – q)}}{{(p + q)}}}}$

$=\frac{\sqrt{p q}}{p}=\sqrt{\frac{q}{p}}$