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, $cot^{2}\theta+tan^{2}\theta=\frac{10}{3}$.

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Updated on: 10-Oct-2022

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