UPSC CDS 2 2023 Math Paper

Concept:Use the double-angle formula for tangent: $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$, and apply it to the geometry of heights and distances.Explanation:Let the angle of elevation at point $R$ be $\theta$.Then the angle of elevation at point $Q$ is $2\theta$.The height of the tower is $h$.$P$ is the foot of the tower.Distance $PQ = x$, distance $QR = y$, so distance $PR = x + y$.From $\text{triangle} PRT$ (right angle at $P$), $\tan \theta = \frac{h}{x + y}$.From $\text{triangle} PQT$ (right angle at $P$), $\tan 2\theta = \frac{h}{x}$.Using the double-angle identity:$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.Substitute the expressions:$\frac{h}{x} = \frac{2 \cdot \frac{h}{x+y}}{1 - \left(\frac{h}{x+y}\right)^2}$.Cancel $h$ (assuming $h \neq 0$):$\frac{1}{x} = \frac{\frac{2}{x+y}}{1 - \frac{h^2}{(x+y)^2}}$.Simplify the denominator:$1 - \frac{h^2}{(x+y)^2} = \frac{(x+y)^2 - h^2}{(x+y)^2}$.Thus, $\frac{1}{x} = \frac{\frac{2}{x+y}}{\frac{(x+y)^2 - h^2}{(x+y)^2}} = \frac{2(x+y)}{(x+y)^2 - h^2}$.Cross-multiply:$(x+y)^2 - h^2 = 2x(x+y)$.Expand left side: $x^2 + 2xy + y^2 - h^2 = 2x^2 + 2xy$.Cancel $2xy$ from both sides: $x^2 + y^2 - h^2 = 2x^2$.Rearrange: $y^2 - h^2 = 2x^2 - x^2 = x^2$.Therefore, $h^2 = y^2 - x^2$.Answer:$h^2 = y^2 - x^2$, which corresponds to option D.