If $\sqrt{3} \cot^{2} \theta - 4 \cot \theta + \sqrt{3} = 0$ then find the value of $\cot^{2} \theta +…
If $\sqrt{3} \cot^{2} \theta - 4 \cot \theta + \sqrt{3} = 0$ then find the value of $\cot^{2} \theta + \tan^{2} \theta$.
Solution
To solve the equation $\sqrt{3} \cot^{2} \theta - 4 \cot \theta + \sqrt{3} = 0$, we first substitute $x = \cot \theta$. This transforms the equation into a quadratic form:
$
\sqrt{3} x^2 - 4x + \sqrt{3} = 0
$
We will use the quadratic formula to find $x$. The quadratic formula is given by:
$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$
Here, $a = \sqrt{3}$, $b = -4$, and $c = \sqrt{3}$. Substituting these values into the quadratic formula gives:
$
x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot \sqrt{3} \cdot \sqrt{3}}}{2\sqrt{3}}
$
Simplifying the expression inside the square root:
$
x = \frac{4 \pm \sqrt{16 - 12}}{2\sqrt{3}} = \frac{4 \pm \sqrt{4}}{2\sqrt{3}} = \frac{4 \pm 2}{2\sqrt{3}}
$
This yields two solutions:
1. $x = \frac{4 + 2}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$
2. $x = \frac{4 - 2}{2\sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$
Thus, $\cot \theta = \sqrt{3}$ or $\cot \theta = \frac{1}{\sqrt{3}}$.
For each case, we need to find $\cot^2 \theta + \tan^2 \theta$. Recall that $\tan \theta = \frac{1}{\cot \theta}$, so:
1. If $\cot \theta = \sqrt{3}$, then $\tan \theta = \frac{1}{\sqrt{3}}$
$
\cot^2 \theta = (\sqrt{3})^2 = 3, \quad \tan^2 \theta = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}
$
$
\cot^2 \theta + \tan^2 \theta = 3 + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3}
$
2. If $\cot \theta = \frac{1}{\sqrt{3}}$, then $\tan \theta = \sqrt{3}$
$
\cot^2 \theta = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}, \quad \tan^2 \theta = (\sqrt{3})^2 = 3
$
$
\cot^2 \theta + \tan^2 \theta = \frac{1}{3} + 3 = \frac{1}{3} + \frac{9}{3} = \frac{10}{3}
$
In both cases, the value of $\cot^2 \theta + \tan^2 \theta$ is $\frac{10}{3}$. Therefore, the correct answer is $\frac{10}{3}$.
Asked in: CBSE
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