If $1+\sin ^{2} \theta=3 \sin 0 \cos 0$, then prove that $\tan 0=1$ or $\frac{1}{2}$
Given, $1+\sin ^{2} A=3 \sin A \cdot \cos A$
On dividing by $\sin ^{2} \theta$ on both sides, we get
$\frac{1}{\sin ^{2} \theta}+1=3 \cdot \cot \theta \quad\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$\Rightarrow \quad \operatorname{cosec}^{2} \theta+1=3 \cdot \cot \theta$ $\left[\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$
$\Rightarrow \quad 1+\cot ^{2} \theta+1=3 \cdot \cot \theta$ $\left[\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right]$
$\Rightarrow \quad \cot ^{2} \theta-3 \cot \theta+2=0$
$\Rightarrow \quad \cot ^{2} \theta-2 \cot \theta-\cot \theta+2=0$ [by splitting the middle term]
$\Rightarrow \cot \theta(\cot \theta-2)-1(\cot \theta-2)=0$
$\Rightarrow \quad(\cot \theta-2)(\cot \theta-1)=0 \Rightarrow \cot \theta=1$ or 2
$\Rightarrow$ $\tan \theta=1$ or $\frac{1}{2}$ $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
Hence proved.
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