Given, $1+\sin ^{2} \theta=3 \sin \theta \cdot \cos \theta$

On dividing by $\sin ^{2} \theta$ on both sides, we get

$\frac{1}{\sin ^{2} \theta}+1=3 \cdot \cot \theta \quad\left[\because \cot P=\frac{\cos \theta}{\sin \theta}\right]$

$\operatorname{cosec}^{2} \theta+1=3 \cdot \cot \theta$ $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$1+\cot ^{2} \theta+1=3 \cdot \cot \theta \quad\left[\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right]$

$\cot ^{2} \theta-3 \cot \theta+2=0$

$\cot ^{2} \theta-2 \cot \theta-\cot \theta+2=0$ [by splitting the middle term]

$\cot \theta(\cot \theta-2)-1(\cot \theta-2)=0$

$(\cot \theta-2)(\cot \theta-1)=0 \Rightarrow \cot \theta=1 \quad$ or $\quad 2$

$\tan \theta=1$ or $\frac{1}{2}$ $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$ Hence proved.