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If $1 + \sin^2 \theta = 3 \sin \theta \cos \theta$, then prove that $\tan \theta = 1$ or $\frac{1}{2}$.
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$1 + \sin^2 \theta = 3 \sin \theta \cos \theta$
$\Rightarrow (\sin^2 \theta + \cos^2\theta) + \sin^2 \theta – 3 \sin \theta \cos \theta = 0$
$\Rightarrow 2 \sin^2 \theta + \cos^2 \theta – 3 \sin \theta \cos \theta = 0$
Dividing by $\cos^2 \theta$, we get
$2 \tan^2 \theta - 3 \tan \theta + 1 = 0$
$\Rightarrow (2 \tan \theta – 1)( \tan \theta – 1) = 0$
$\therefore \tan \theta = \frac{1}{2}$ or $1$
$\Rightarrow (\sin^2 \theta + \cos^2\theta) + \sin^2 \theta – 3 \sin \theta \cos \theta = 0$
$\Rightarrow 2 \sin^2 \theta + \cos^2 \theta – 3 \sin \theta \cos \theta = 0$
Dividing by $\cos^2 \theta$, we get
$2 \tan^2 \theta - 3 \tan \theta + 1 = 0$
$\Rightarrow (2 \tan \theta – 1)( \tan \theta – 1) = 0$
$\therefore \tan \theta = \frac{1}{2}$ or $1$