If tan⁡α=Pq where α=2β, α being an acute angle then 12[pcosec⁡2β+qsec⁡2β] is equal to

12[pcosec⁡2β+qsec⁡2β]⇒p2+q22pp2+q2cosec⁡2β+qp2+q2sec⁡2β=p2+q22sinαsin2β+cosαcos2β=p2+q222=p2+q2

If tan⁡α=Pq where α=2β, α being an acute angle then 12[pcosec⁡2β+qsec⁡2β] is equal to

12[pcosec⁡2β+qsec⁡2β]⇒p2+q22pp2+q2cosec⁡2β+qp2+q2sec⁡2β=p2+q22sinαsin2β+cosαcos2β=p2+q222=p2+q2

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If $\tan α = \frac{P}{q} where α = 2 β, α$ being an acute angle then $\frac{1}{2} [pcosec 2 β + qsec 2 β]$ is equal to

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$\sqrt{p^{2} + q^{2}}$

$\sqrt{p^{2} - q^{2}}$

$\sqrt{2 p^{2} + q^{2}}$

$2 \sqrt{p^{2} + q^{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} [pcosec 2 β + qsec 2 β] \\ \Rightarrow \frac{\sqrt{p^{2} + q^{2}}}{2} [\frac{p}{\sqrt{p^{2} + q^{2}}} cosec 2 β + \frac{q}{\sqrt{p^{2} + q^{2}}} \sec 2 β] \\ = \frac{\sqrt{p^{2} + q^{2}}}{2} [\frac{\sin α}{\sin 2 β} + \frac{\cos α}{\cos 2 β}] = \frac{\sqrt{p^{2} + q^{2}}}{2} (2) = \sqrt{p^{2} + q^{2}} \end{array}$