# If $\tan\alpha=\large\frac{p}{q}$ where $\alpha=6\beta$ .$\alpha$ being acute angle,then $\large\frac{1}{2}$$[p cosec 2\beta-q\sec 2\beta] is equal to \begin{array}{1 1}(a)\;\sqrt{p^2+q^2}&(b)\;\sqrt{p^2-q^2}\\(c)\;\sqrt{2p^2+q^2}&(d)\;None\;of\;these\end{array} ## 1 Answer we have \tan\alpha =\large\frac{p}{q} \therefore \sin\alpha=\large\frac{p}{\sqrt{p^2+q^2}} \cos\alpha=\large\frac{q}{\sqrt{p^2+q^2}} Now \large\frac{1}{2}$$[p cosec 2\beta-q\sec 2\beta]$
$\Rightarrow \large\frac{\sqrt{p^2+q^2}}{2}\bigg[\large\frac{p}{\sqrt{p^2+q^2}}$$cosec 2\beta-\large\frac{q}{\sqrt{p^2+q^2}}$$\sec 2\beta\bigg]$
$\Rightarrow \large\frac{\sqrt{p^2+q^2}}{2}\bigg[\large\frac{\sin \alpha}{\sin 2\beta}-\frac{\cos \alpha}{\cos 2\beta}\bigg]$
$\Rightarrow \sqrt{p^2+q^2}\bigg[\large\frac{\sin(\alpha-2\beta)}{2\sin 2\beta\cos 2\beta}\bigg]$
$\alpha=6\beta$
$2\sin 2\beta\cos 2\beta=\sin 4\beta$
$\Rightarrow \sqrt{p^2+q^2}\bigg[\large\frac{\sin(6\beta-2\beta)}{\sin 4\beta}\bigg]$
$\Rightarrow \sqrt{p^2+q^2}\bigg[\large\frac{\sin(4\beta)}{\sin 4\beta}\bigg]$
$\Rightarrow \sqrt{p^2+q^2}$
Hence (a) is the correct option.