Question : If $\sqrt{3} \tan ^2 \theta-4 \tan \theta+\sqrt{3}=0$, then what is the value of $\tan ^2 \theta+\cot ^2 \theta$?
Option 1: $\frac{4}{3}$
Option 2: $\frac{10}{3}$
Option 3: $3$
Option 4: $\frac{6}{5}$
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Correct Answer: $\frac{10}{3}$
Solution : $\sqrt{3} \tan ^2 \theta-4 \tan \theta+\sqrt{3}=0$ Multiplying by $\sqrt{3}$ on both sides, we get, $3\tan^2\theta−4\sqrt{3}\tan \theta+3=0$ ⇒ $3\tan^2\theta−3\sqrt{3}\tan\theta−\sqrt{3}\tan\theta+3=0$ ⇒ $3\tan\theta(\tan\theta−\sqrt{3})−\sqrt{3}(\tan\theta−\sqrt{3})=0$ ⇒ $(3\tan\theta−\sqrt{3})(\tan\theta−\sqrt{3})=0$ ⇒ $\tan\theta=\frac{1}{\sqrt{3}}$ or, $\tan\theta=\sqrt{3}$ ⇒ $\theta=30°$ or, $\theta=60°$ So, $\tan ^2 \theta+\cot ^2 \theta$ $=\tan^2 30° + \cot^2 30°$ $=\frac{1}{3}+3$ $=\frac{10}{3}$ Hence, the correct answer is $\frac{10}{3}$.
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