If $1 + \sin^2\theta = 3 \sin \theta \cos \theta$, then $\tan \theta$ can take values

Given: $1 + \sin^2\theta = 3 \sin \theta \cos \theta$

$\sin^2\theta + \cos^2\theta = 1$

$\sin^2\theta = 1 - \cos^2\theta$

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$1 + (1 - \cos^2\theta) = 3 \sin \theta \cos \theta$

$2 - \cos^2\theta = 3 \sin \theta \cos \theta$

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$\dfrac{2 - \cos^2\theta}{\cos^2\theta} = \dfrac{3 \sin \theta \cos \theta}{\cos^2\theta}$

$\dfrac{2}{\cos^2\theta} - 1 = 3 \cdot \dfrac{\sin\theta}{\cos\theta}$

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$\dfrac{1}{\cos^2\theta} = 1 + \tan^2\theta$ and $\dfrac{\sin\theta}{\cos\theta} = \tan\theta$

$2(1 + \tan^2\theta) - 1 = 3\tan\theta$

$2 + 2\tan^2\theta - 1 = 3\tan\theta$

$1 + 2\tan^2\theta = 3\tan\theta$

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$2\tan^2\theta - 3\tan\theta + 1 = 0$

$t = \tan\theta$

$2t^2 - 3t + 1 = 0$

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$t = \dfrac{3 \pm \sqrt{9 - 8}}{4}$

$t = \dfrac{3 \pm 1}{4}$

$t = \dfrac{4}{4} = 1$ or $t = \dfrac{2}{4} = \dfrac{1}{2}$

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Therefore, $\tan \theta$ can take the values $\dfrac{1}{2}$ and $1$.