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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