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Having trouble solving this trigonometry problem I came across. Any help would be greatly appreciated

If $\sqrt{3}cot^2\theta -4cot\theta + \sqrt{3} = 0 $

Find the value of $cot^2\theta + tan^2\theta$

Thank you so much guys!

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  • $\begingroup$ @Abhishekstudent I got the answer. Given in a nice and concise proof below. Thanks anyway! $\endgroup$ – Aman Bhargava Sep 13 '16 at 12:32
  • $\begingroup$ Good to know! Best Of Luck! $\endgroup$ – Abhishekstudent Sep 13 '16 at 12:33
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Hint:-

Let,$\cot\theta=x$

So, $\sqrt3 x^2-4x+\sqrt3=0$

$\implies\sqrt3x^2-3x-x+\sqrt3=0$

$\implies(\sqrt3x-1)(x-\sqrt3)=0$

So,$\cot\theta=\color{red}{\frac{1}{\sqrt3}}$ or $\color{blue}{\sqrt3}$

So,$\tan\theta=\color{red}{\sqrt3}$ or $\color{blue}{\frac{1}{\sqrt3}}$

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  • $\begingroup$ Okay wow, that was really fast. Thanks so much! I can see what I was missing. It turned out to be a simple quadratic! But thank you so much! I love this community :') $\endgroup$ – Aman Bhargava Sep 13 '16 at 12:26
  • $\begingroup$ @AmanBhargava Welcome to Math SE!!If you like my answer you can thank me by upvoting and accepting the answer!! $\endgroup$ – tatan Sep 13 '16 at 12:27
  • $\begingroup$ I surely will! It doesn't allow me to do so within 10 mins, but I will surely upvote and accept your answer. Nice to know that this isn't a hostile community like Stackoverflow. People here are really friendly and helpful! $\endgroup$ – Aman Bhargava Sep 13 '16 at 12:29
  • $\begingroup$ @AmanBhargava Thanks!!You see,I also had a lot of problem when I began in StackOverflow....its not that they are hostile but StackOverflow is mostly for high end experts so beginners like we are have problem in adapting at first.... $\endgroup$ – tatan Sep 13 '16 at 12:31
  • $\begingroup$ @AmanBhargava 10th Grade (I'm from India)... $\endgroup$ – tatan Sep 13 '16 at 12:35
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$$4 \cot \theta = \sqrt{3}(\cot^2 \theta + 1) = \sqrt{3}\csc^2\theta$$ Simplifying, $$ 4 \cos\theta = \sqrt{3} \sin \theta \csc^2\theta = \sqrt{3} \csc \theta $$ Again, simplifying, $$ 4 \cos \theta \sin \theta = \sqrt{3} \implies \sin 2\theta = \frac{\sqrt{3}}{2} $$ Two results come out: $$ 2\theta = \frac{(2n+1)\pi}{3} \implies \theta = \frac{(2n+1)\pi}{6} $$ or:$$ \implies 2\theta = -\frac{2n\pi}{3} \implies \theta = -\frac{n\pi}{3} $$

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