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# If $2 - {\cos ^2}\theta = 3\sin \theta \cos \theta \ne \cos \theta$ than find the value of $\cot \theta$A. $\dfrac{1}{2}$B. $0$C. $- 1$D. $2$

Last updated date: 13th Jun 2024
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Hint: Here we will proceed by converting the given equation in terms of $\tan$ by using the formulae of trigonometric ratios and trigonometric identities. Then solve the obtained equation by grouping the common terms. Further convert tan to cot to get the required answer.

The equation is $2 - {\cos ^2}\theta = 3\sin \theta \cos \theta$.
Dividing both sides with ${\cos ^2}\theta$, we have
$\Rightarrow \dfrac{{2 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = \dfrac{{3\sin \theta \cos \theta }}{{{{\cos }^2}\theta }} \\ \Rightarrow \dfrac{2}{{{{\cos }^2}\theta }} - \dfrac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = \dfrac{{3\sin \theta }}{{\cos \theta }} \\ \Rightarrow 2{\sec ^2}\theta - 1 = 3\tan \theta {\text{ }}\left[ {\because \dfrac{1}{{{{\cos }^2}\theta }} = {{\sec }^2}\theta ,\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \\$

We know that ${\sec ^2}\theta = {\tan ^2}\theta + 1$. By substituting this formula, we have
$\Rightarrow 2\left( {{{\tan }^2}\theta + 1} \right) - 1 = 3\tan \theta \\ \Rightarrow 2{\tan ^2}\theta + 2 - 1 = 3\tan \theta \\ \Rightarrow 2{\tan ^2}\theta - 3\tan \theta + 1 = 0 \\$
Splitting and grouping the common terms, we have
$\Rightarrow 2{\tan ^2}\theta - 2\tan \theta - \tan \theta + 1 = 0 \\ \Rightarrow 2\tan \theta \left( {\tan \theta - 1} \right) - 1\left( {\tan \theta - 1} \right) = 0 \\ \Rightarrow \left( {2\tan \theta - 1} \right)\left( {\tan \theta - 1} \right) = 0 \\ \therefore \tan \theta = \dfrac{1}{2},1 \\$
We know that $\cot \theta = \dfrac{1}{{\tan \theta }}$. So, we have
$\therefore \cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{1}{{\dfrac{1}{2}}} = 2 \\ {\text{ }}\cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{1}{1} = 1 \\$
Therefore, the values of $\cot \theta$ are 1 and 2.
Thus, the correct answer is D. 2

So, the correct answer is “Option D”.

Note: Here we have used the formulae of trigonometric ratios of $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta ,\dfrac{1}{{{{\cos }^2}\theta }} = {\sec ^2}\theta ,\dfrac{1}{{\tan \theta }} = \cot \theta$. And the trigonometric identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta$ to solve the given problem.