Correct Answer: $60^{\circ}$

Solution : Given: The value of $\frac{\cos ^2 \theta}{\cot ^2 \theta–\cos ^2 \theta}=3$, where $0^{\circ}<\theta<90^{\circ}$.
Use the trigonometric identity, $\sin^2 \theta+\cos^2 \theta=1$.
$\frac{\cos ^2 \theta}{\cot ^2 \theta–\cos ^2 \theta}=3$
⇒ $\cos ^2 \theta=3(\cot ^2 \theta–\cos ^2 \theta)$
⇒ $\cos ^2 \theta=3(\frac{\cos ^2 \theta}{\sin^2 \theta}–\cos ^2 \theta)$
Dividing both sides by $\cos^2 \theta$,
⇒ $1=3(\frac{1}{\sin ^2 \theta}–1)$
⇒ $1=3(\frac{1–\sin ^2 \theta}{\sin^2 \theta})$
⇒ $\sin^2 \theta=3–3\sin ^2 \theta$
⇒ $4\sin^2 \theta=3$
⇒ $\sin^2 \theta=\frac{3}{4}$
⇒ $\sin \theta=\sqrt{\frac{3}{4}}$
⇒ $\sin \theta=\frac{\sqrt3}{2}$
⇒ $\sin \theta=\sin 60^{\circ}$
⇒ $\theta= 60^{\circ}$
Hence, the correct answer is $60^{\circ}$.