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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If 1 + sin θ + ^ 2 θ + ^ 3 θ +... ∞ = 4 + 2 3 , 0 < θ < π, then

If 1+sinθ+sin⁡2θ+sin⁡3θ+...∞=4+23,0<θ<π,1 + sin \theta + \sin^{2} \theta + \sin^{3} \theta +... \infty = 4 + 2\sqrt{3}, 0 < \theta < \pi, then

Options

  1. Aθ=π3\theta = \dfrac{\pi}{3}
  2. Bθ=π6\theta = \dfrac{\pi}{6}
  3. Cθ=π3\theta = \dfrac{\pi}{3} or π6\dfrac{\pi}{6}
  4. Dθ=π3\theta = \dfrac{\pi}{3} or 2π3\dfrac{2\pi}{3}

Correct answer

D. θ=π3\theta = \dfrac{\pi}{3} or 2π3\dfrac{2\pi}{3}

Step-by-step solution

Given, 1+sinθ+sin⁡2θ+sin⁡3θ+…∞=4+231 + sin \theta + \sin^{2} \theta + \sin^{3} \theta + \ldots \infty = 4 + 2\sqrt{3}

⇒11−sinθ=4+23[∵0<sinθ<1]⇒ \dfrac{1}{1 - sin \theta} = 4 + 2\sqrt{3} [ ∵ 0 < sin \theta < 1 ]

⇒1−sinθ=4−2316−12=1−32⇒ 1 - sin \theta = \dfrac{4 - 2\sqrt{3}}{16 - 12} = 1 - \dfrac{\sqrt{3}}{2}

⇒sinθ=32⇒ sin \theta = \dfrac{\sqrt{3}}{2}

⇒θ=π3⇒ \theta = \dfrac{\pi}{3} or 2π3\dfrac{2\pi}{3}
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