Answer
414.9k+ views
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.
Complete step-by-step 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.
Complete step-by-step 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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)