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If $\sqrt{3}cot^{2}\theta-4cot\theta+\sqrt{3}=0$, then find the value of $cot^{2}\theta+tan^{2}\theta$.
Given: $\sqrt{3}cot^{2}\theta-4cot\theta+\sqrt{3}=0$.
To do: To find the value of $cot^{2}\theta+tan^{2}\theta$.
Solution:
$\sqrt{3}cot^{2}\theta-4cot\theta+\sqrt{3}=0$
$\Rightarrow \sqrt{3}cot^{2}\theta-3cot\theta-cot\theta+\sqrt{3}=0$
$\Rightarrow \sqrt{3} cot\theta( cot\theta-\sqrt{3})-( cot\theta-\sqrt{3})=0$
$\Rightarrow ( \sqrt{3}cot\theta-1)(cot\theta-\sqrt{3})=0$
If $\sqrt{3}cot\theta-1=0$
$\Rightarrow cot\theta=\frac{1}{\sqrt{3}}$
If $cot\theta-\sqrt{3}=0$
$\Rightarrow cot\theta=\sqrt{3}$
If $cot\theta=\frac{1}{\sqrt{3}}$, then
$cot^{2}\theta+tan^{2}\theta=cot^{2}\theta+\frac{1}{cot^{2}\theta}$
$=( \frac{1}{\sqrt{3}})^{2}+( \frac{1}{\frac{1}{\sqrt{3}}})^{2}$
$=\frac{1}{3}+3$
$=\frac{10}{3}$
If $cot\theta=\sqrt{3}$
$\Rightarrow cot^{2}\theta+tan^{2}\theta=cot^{2}\theta+\frac{1}{cot^{2}\theta}$
$=( \sqrt{3})^{2}+( \frac{1}{\sqrt{3}})^{2}$
$=3+\frac{1}{3}$
$=\frac{10}{3}$
Thus, $cot^{2}\theta+tan^{2}\theta=\frac{10}{3}$.