UPSC NDA I and II 2020 Math Solved Paper

Concept:Given $p \tan x = q \tan y$ and expressions for $\tan^2 x$ and $\tan^2 y$, we need $\frac{p^2}{q^2}$ using ratio of squares of tangents.Explanation:From $p \tan x = q \tan y$, we have $\frac{p}{q} = \frac{\tan y}{\tan x}$.Square both sides: $\frac{p^2}{q^2} = \frac{\tan^2 y}{\tan^2 x}$.Use provided values: $\tan^2 y = \frac{d-a}{b-d}$ and $\tan^2 x = \frac{c-b}{a-c}$.Thus $\frac{p^2}{q^2} = \frac{\frac{d-a}{b-d}}{\frac{c-b}{a-c}} = \frac{(d-a)(a-c)}{(b-d)(c-b)}$.Simplify signs: $(d-a) = -(a-d)$, $(a-c) = -(c-a)$, $(b-d) = -(d-b)$, $(c-b) = -(b-c)$.The product of two negatives is positive, and the denominator also has two negatives: $\frac{(-(a-d))(-(c-a))}{(-(d-b))(-(b-c))} = \frac{(a-d)(c-a)}{(d-b)(b-c)}$.Rewriting denominator in order: $(d-b) = -(b-d)$, but we already accounted signs. Alternatively, $\frac{(a-d)(c-a)}{(b-c)(d-b)}$ matches option B after checking signs.Thus $\frac{p^2}{q^2} = \frac{(a-d)(c-a)}{(b-c)(d-b)}$.Answer:$\frac{p^2}{q^2} = \frac{(a-d)(c-a)}{(b-c)(d-b)}$, which is option B.