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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If ^ 3 θ - ^ 3 θ sin θ - cos θ - cos θ 1 + ^ 2 θ - 2tan θ cot θ = -...

If sin⁡3θ−cos⁡3θsinθ−cosθ−cosθ1+cot⁡2θ−2tanθcotθ=−1(∀θ∈[0,2π]),\dfrac{\sin^{3} \theta - \cos^{3} \theta}{sin \theta - cos \theta} - \dfrac{cos \theta}{1 + \cot^{2} \theta} - 2tan \theta cot \theta = - 1 \left(\forall \theta \in \left[0, 2\pi\right]\right), then

Options

  1. Aθ∈(0,π2)−{π4}\theta \in \left(0, \dfrac{\pi}{2}\right) - \left\{\dfrac{\pi}{4}\right\}
  2. Bθ∈(π2,π)−{3π4}\theta \in \left(\dfrac{\pi}{2}, \pi\right) - \left\{\dfrac{3\pi}{4}\right\}
  3. Cθ∈(π,3π2)−{5π4}\theta \in \left(\pi, \dfrac{3\pi}{2}\right) - \left\{\dfrac{5\pi}{4}\right\}
  4. Dθ∈(0,π)−{π4,π2}\theta \in \left(0, \pi\right) - \left\{\dfrac{\pi}{4}, \dfrac{\pi}{2}\right\}

Correct answer

D. θ∈(0,π)−{π4,π2}\theta \in \left(0, \pi\right) - \left\{\dfrac{\pi}{4}, \dfrac{\pi}{2}\right\}

Step-by-step solution

Since,& sinθ−cosθ≠0⇒tanθ≠1sin \theta - cos \theta \ne 0 ⇒ tan \theta \ne 1

To define cotθ,θ≠0,π,2πcot \theta, \theta \ne 0, \pi, 2\pi and

to define tanθ,θ≠π2,3π2tan \theta, \theta \ne \dfrac{\pi}{2}, \dfrac{3\pi}{2}
⇒θ≠π4,5π4,π2,3π2,0,π,2π⇒ \theta \ne \dfrac{\pi}{4}, \dfrac{5\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}, 0, \pi, 2\pi
Now, sin⁡2θ+cos⁡2θ+sinθcosθ−∣sin⁡θ∣cosθ−2=−1\sin^{2} \theta + \cos^{2} \theta + sin \theta cos \theta - \left|\sin \theta\right|cos \theta - 2 = - 1
⇒1+sinθcosθ−∣sin⁡θ∣cosθ=1⇒ 1 + sin \theta cos \theta - \left|\sin \theta\right|cos \theta = 1
⇒cosθ(sinθ−∣sin⁡θ∣)=0⇒ cos \theta\left(sin \theta - \left|\sin \theta\right|\right) = 0
⇒cosθ=0⇒ cos \theta = 0 or sin \theta = \left|sin \theta\right|<br />⇒ \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2} or \theta = \left[0, \pi\right]<br />⇒ θ∈(0,π)−{π4,π2}\theta \in \left(0, \pi\right) - \left\{\dfrac{π}{4}, \dfrac{π}{2}\right\}

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