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Question: If \(2 - {\cos ^2}\theta = 3\sin \theta \cos \theta \ne \cos \theta \) than find the value of \(\cot...

If 2−cos⁡2θ=3sin⁡θcos⁡θ≠cos⁡θ2 - {\cos ^2}\theta = 3\sin \theta \cos \theta \ne \cos \theta than find the value of cot⁡θ\cot \theta
A. 12\dfrac{1}{2}
B. 00
C. −1 - 1
D. 22

Explanation

Solution

Here we will proceed by converting the given equation in terms of tan⁡\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⁡2θ=3sin⁡θcos⁡θ2 - {\cos ^2}\theta = 3\sin \theta \cos \theta .
Dividing both sides with cos⁡2θ{\cos ^2}\theta , we have

⇒2−cos⁡2θcos⁡2θ=3sin⁡θcos⁡θcos⁡2θ ⇒2cos⁡2θ−cos⁡2θcos⁡2θ=3sin⁡θcos⁡θ ⇒2sec⁡2θ−1=3tan⁡θ [∵1cos⁡2θ=sec⁡2θ,sin⁡θcos⁡θ=tan⁡θ]  \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⁡2θ=tan⁡2θ+1{\sec ^2}\theta = {\tan ^2}\theta + 1. By substituting this formula, we have

⇒2(tan⁡2θ+1)−1=3tan⁡θ ⇒2tan⁡2θ+2−1=3tan⁡θ ⇒2tan⁡2θ−3tan⁡θ+1=0  \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

⇒2tan⁡2θ−2tan⁡θ−tan⁡θ+1=0 ⇒2tan⁡θ(tan⁡θ−1)−1(tan⁡θ−1)=0 ⇒(2tan⁡θ−1)(tan⁡θ−1)=0 ∴tan⁡θ=12,1  \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⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}. So, we have

∴cot⁡θ=1tan⁡θ=112=2  cot⁡θ=1tan⁡θ=11=1  \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⁡θ\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 sin⁡θcos⁡θ=tan⁡θ,1cos⁡2θ=sec⁡2θ,1tan⁡θ=cot⁡θ\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⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta to solve the given problem.