If \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) denote the lengths of the perpendiculars from the origin on the lines \(\mathrm{xsec} \alpha+\mathrm{ycosec} \alpha=2 \mathrm{a}\) and \(x \cos \alpha+y \sin \alpha=a \cos 2 \alpha\) respectively then \(\left[\left(\mathrm{P}_{1} / \mathrm{P}_{2}\right)+\left(\mathrm{P}_{2} / \mathrm{P}_{1}\right)\right]^{2}\) is equal to \(\ldots \ldots\) (a) \(4 \sin ^{2} 4 \alpha\). (b) \(4 \cos ^{2} 4 \alpha\) (c) \(4 \operatorname{cosec}^{2} 4 \alpha\) (d) \(4 \sec ^{2} 4 \alpha\)

The given lines are: \(x\sec\alpha + y\csc\alpha = 2a\) \(x\cos\alpha + y\sin\alpha = a\cos{2\alpha}\)

The formula for the distance between a point (x₀, y₀) and a line Ax + By + C = 0 is given as: \(P = \frac{|Ax₀ + By₀ + C|}{\sqrt{A^2 + B^2}}\) As we are considering the perpendiculars from the origin, the point (x₀, y₀) is (0, 0).

For line 1: \(P_1 = \frac{|2a|}{\sqrt{(\sec^2\alpha + \csc^2\alpha)}}\) For line 2: \(P_2 = \frac{|a\cos{2\alpha}|}{\sqrt{(\cos^2\alpha + \sin^2\alpha)}}\)

First, let's simplify the denominators using trigonometric identities: \(\sec^2\alpha = 1 + \tan^2\alpha\) \(\csc^2\alpha = 1 + \cot^2\alpha\) \(\cos^2\alpha + \sin^2\alpha = 1\) Now we have: \(P_1 = \frac{2a}{\sqrt{1+\tan^2\alpha + 1+\cot^2\alpha}}\) \(P_2 = a\cos{2\alpha}\)

The expression is given as: \(\left[\frac{P_1}{P_2} + \frac{P_2}{P_1}\right]^2\) Substitute the values of P1 and P2: \(\left[\frac{\frac{2a}{\sqrt{1+\tan^2\alpha + 1+\cot^2\alpha}}}{a\cos{2\alpha}} + \frac{a\cos{2\alpha}}{\frac{2a}{\sqrt{1+\tan^2\alpha + 1+\cot^2\alpha}}}\right]^2\)

Cancel out the common factors a and simplify the expression: \(\left[\frac{2}{\cos{2\alpha}\sqrt{2+\tan^2\alpha + \cot^2\alpha}} + \frac{\cos{2\alpha}\sqrt{2+\tan^2\alpha + \cot^2\alpha}}{2}\right]^2\) We notice that we have a sum of conjugates, so square the expression to eliminate the square roots: \(\left[\frac{2}{\cos{2\alpha}\sqrt{2+\tan^2\alpha + \cot^2\alpha}}\right]^2+ 2 + \left[\frac{\cos{2\alpha}\sqrt{2+\tan^2\alpha + \cot^2\alpha}}{2}\right]^2\)

Simplify the expression: \(\frac{4}{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}+2+\frac{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}{4}\) Multiply the last term by 4/4 to unify the denominators: \(\frac{4}{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}+2+\frac{4\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}{16}\) Add the terms: \(\frac{4+8\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)+4\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}\) Simplify the numerator: \(\frac{4\left(1+\cos^2{2\alpha}\right)\left(2+\tan^2\alpha+\cot^2\alpha\right)}{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}\) We know that \(1+\cos^2{2\alpha}=\sec^2{4\alpha}\) , So the expression becomes: \(\frac{4\sec^2{4\alpha}(2+\tan^2\alpha+\cot^2\alpha)}{\cos^2{2\alpha}(2+\tan^2\alpha+\cot^2\alpha)}\) Now, cancel the common terms in the numerator and denominator: \(\frac{4\sec^2{4\alpha}}{\cos^2{2\alpha}}\) This is equal to the given option (d). Therefore, the expression equals \(4\sec^2{4\alpha}\).