If $1+\sin ^{2} \theta=3 \sin \theta \cos \theta,$ then prove that $\tan \theta=1$ or $\frac{1}{2}$.

(A) Given equation: $1+\sin ^{2} \theta=3 \sin \theta \cos \theta$
Divide both sides by $\cos^{2} \theta$:
$\frac{1}{\cos^{2} \theta} + \frac{\sin^{2} \theta}{\cos^{2} \theta} = \frac{3 \sin \theta \cos \theta}{\cos^{2} \theta}$
Using the identities $\sec^{2} \theta = \frac{1}{\cos^{2} \theta}$,$\tan \theta = \frac{\sin \theta}{\cos \theta}$,and $\sec^{2} \theta = 1 + \tan^{2} \theta$:
$(1 + \tan^{2} \theta) + \tan^{2} \theta = 3 \tan \theta$
Rearranging the terms to form a quadratic equation:
$2 \tan^{2} \theta - 3 \tan \theta + 1 = 0$
Factorizing the quadratic equation:
$2 \tan^{2} \theta - 2 \tan \theta - \tan \theta + 1 = 0$
$2 \tan \theta (\tan \theta - 1) - 1 (\tan \theta - 1) = 0$
$(2 \tan \theta - 1) (\tan \theta - 1) = 0$
Therefore,$\tan \theta - 1 = 0$ or $2 \tan \theta - 1 = 0$
$\tan \theta = 1$ or $\tan \theta = \frac{1}{2}$.
Hence proved.