Correct Answer: $\frac{\cot ^2 \alpha+\cot ^2 \beta}{\cot ^2 \gamma+\cot ^2 \delta}$

Solution :
Let K be the point on the top of the tower and the height of the clock tower $OK$ be $h$ cm
$OK$ is perpendicular to $PR$ and $SQ$.
In Δ$POK$,
$\tan α = \frac{OK}{OP}$
⇒ $OP = \frac{h}{\tan α}$
⇒ $OP = h \cot α$
Similarly, In ΔQOK
$\tan β = \frac{OK}{OQ}$
⇒ $OQ = \frac{h}{\tan β}$
⇒ $OQ = h \cot β$
In $ΔPOQ$ formed by joining $P$ and $Q$, $OP$ is perpendicular to $OQ$, and then
⇒ $PQ^2 = OP^2 + OQ^2$
⇒ $PQ^2 = h^2 \cot^2 α + h^2 \cot^2 β$
⇒ $PQ^2 = h^2 (\cot^2 α + \cot^2 β)$
Similarly,
$RS^2 = h^2 (\cot^2 γ + \cot^2 δ)$
⇒ $(\frac{PQ}{RS})^2 = \frac{h^2 (\cot^2 α + \cot^2 β)}{h^2 (\cot^2 γ + \cot^2 δ)}$
⇒ $(\frac{PQ}{RS})^2 = \frac{\cot^2 α + \cot^2 β}{\cot^2 γ + \cot^2 δ}$
Hence, the correct answer is $\frac{\cot^2 α + \cot^2 β}{\cot^2 γ + \cot^2 δ}$.