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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)/(6) sinθ , cosθ and tan θ are in geometric progression, then...

If 16sinθ\dfrac{1}{6} sin\theta, cosθcos\theta and tanθtan \theta are in geometric progression, then the complete solution set of θ\theta is

Options

  1. A{θ:θ=2nπ±(π6),n∈I}\left\{\theta: \theta = 2n\pi \pm \left(\dfrac{π}{6}\right), n \in I\right\}
  2. B{θ:θ=2nπ±(π3),n\inI}\left\{\theta: \theta = 2n\pi \pm \left(\dfrac{\pi}{3}\right), n\inI\right\}
  3. C{θ:θ=nπ+(−1)n(π3),n\inI}\left\{\theta: \theta = n\pi + {\left(- 1\right)}^{n}\left(\dfrac{\pi}{3}\right), n\inI\right\}
  4. D{θ:θ=nπ+π3,n\inI}\left\{\theta: \theta = n\pi + \dfrac{π}{3}, n\inI\right\}

Correct answer

B. {θ:θ=2nπ±(π3),n\inI}\left\{\theta: \theta = 2n\pi \pm \left(\dfrac{\pi}{3}\right), n\inI\right\}

Step-by-step solution

\cos^{2}\theta = \left(\ddfrac{1}{6}sin\theta\right)\llleft(tanθ)</p><p>lleft(tan\theta)</p><p>⇒ \cos^{2} θ=16(sin⁡2θcosθ)<br/><br/>\theta = \dfrac{1}{6} \left(\dfrac{\sin^{2} \theta}{cos \theta}\right)<br /><br />⇒6\cos^{3} \theta =1 - \cos^{2} \theta<br/><br/><br /><br />⇒6\cos^{3} \theta + \cos^{2} \theta -1 = 0<br/><br/>⇒ <br /><br />⇒~cos \theta = 12 <br/><br/>\dfrac{1}{2}~<br /><br />⇒ \theta =2n\pi \pm π3,\dfrac{π}{3},n \in I .

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