If $\tan\alpha=\large\frac{p}{q}$ where $\alpha=6\beta$ .$\alpha$ being acute angle,then $\large\frac{1}{2}$$[p\; \mathrm{ cosec} 2\beta-q\sec 2\beta]$ is equal to

Question

$\begin{array}{1 1}(a)\;\sqrt{p^2+q^2}&(b)\;\sqrt{p^2-q^2}\\(c)\;\sqrt{2p^2+q^2}&(d)\;None\;of\;these\end{array}$