If \(p_{1}\) and \(p_{2}\) are the lengths of perendiculars from the origin to the lines
\(x \sin \theta+y \cos \theta=5 \cos 2 \theta\) and \(x \operatorname{cosec} \theta+y \sec \theta-5=0\) respectively, then
\(p_{1}^{2}+4 p_{2}^{2}=\)

As per condition given, we write
\(\mathrm{p}_{1} =\frac{|-5 \cos 2 \theta|}{\sqrt{\sin ^{2} \theta+\cos ^{2} \theta}} \text { and } p_{2}=\frac{|-5|}{\sqrt{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}} \)
\( \therefore \mathrm{p}_{1}^{2}=\frac{25 \cos ^{2} 2 \theta}{1} \text { and } 4 \mathrm{p}_{2}^{2} =\frac{4(25)}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta} \)
\( \therefore \mathrm{p}_{1}^{2}+4 \mathrm{p}_{2}^{2} =25\left(\cos ^{2} \theta-\sin ^{2} \theta\right)^{2}+\frac{100}{\left(\frac{1}{\sin ^{2} \theta}\right)+\frac{1}{\cos ^{2} \theta}} \)
\( =25(\cos \theta-\sin \theta)^{2}(\cos \theta+\sin \theta)^{2}+100 \sin ^{2} \theta \cos ^{2} \theta \)
\( =25(1-\sin 2 \theta)(1+\sin 2 \theta)+25\left(4 \sin ^{2} \theta \cos ^{2} \theta\right) \)
\( =25\left(1-\sin 2 \theta+\sin 2 \theta-\sin ^{2} 2 \theta\right)+25(2 \sin \theta \cos \theta)^{2} \)
\( =25\left(1-\sin ^{2} 2 \theta\right)+25(\sin 2 \theta)^{2} \)
\( =25\left(\cos ^{2} 2 \theta\right)+25\left(\sin ^{2} 2 \theta\right)=25\)