Correct Answer: $\frac{13}{12}$

Solution : Given: $4\sin^{2}\theta-1=0$
$⇒4\sin^{2}\theta-1=0$
$⇒4\sin^{2}\theta=1$
$⇒\sin^{2}\theta=\frac{1}{4}$
$⇒\sin\theta=\pm\frac{1}{2}$
Since we're given that $0^{\circ}< \theta< 90^{\circ}$,
So, $\sin\theta$ is positive in this range.
$⇒\sin\theta=\frac{1}{2}=\sin30^\circ$
$\therefore \theta = 30^\circ$
Putting the value of $\theta$, we get,
$\therefore\cos^{2}\theta+\tan^{2}\theta=\cos^{2}30^\circ+\tan^{2}30^\circ=(\frac{\sqrt3}{2})^2+(\frac{1}{\sqrt3})^2=\frac{3}{4}+\frac{1}{3}=\frac{9+4}{12}=\frac{13}{12}$
Hence, the correct answer is $\frac{13}{12}$.