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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If cosec θ - 3 θ = 4 and θ ∈ (0, (π)/(2)) , then the value of θ is :

If cosec⁡θ−3sec⁡θ=4\operatorname{cosec} \theta - \sqrt{3} \sec \theta = 4 and θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right), then the value of θ\theta is :

Options

  1. Aπ12\dfrac{\pi}{12}
  2. Bπ3\dfrac{\pi}{3}
  3. Cπ18\dfrac{\pi}{18}
  4. Dπ30\dfrac{\pi}{30}

Correct answer

C. π18\dfrac{\pi}{18}

Step-by-step solution

The given equation is cosec⁡θ−3sec⁡θ=4\operatorname{cosec} \theta - \sqrt{3} \sec \theta = 4.

Converting to sine and cosine, we get:
1sin⁡θ−3cos⁡θ=4\dfrac{1}{\sin \theta} - \dfrac{\sqrt{3}}{\cos \theta} = 4

⇒cos⁡θ−3sin⁡θsin⁡θcos⁡θ=4\Rightarrow \dfrac{\cos \theta - \sqrt{3} \sin \theta}{\sin \theta \cos \theta} = 4

⇒cos⁡θ−3sin⁡θ=4sin⁡θcos⁡θ\Rightarrow \cos \theta - \sqrt{3} \sin \theta = 4 \sin \theta \cos \theta

Dividing both sides by 22:
12cos⁡θ−32sin⁡θ=2sin⁡θcos⁡θ\dfrac{1}{2} \cos \theta - \dfrac{\sqrt{3}}{2} \sin \theta = 2 \sin \theta \cos \theta

Using the identities sin⁡(π6)=12\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}, cos⁡(π6)=32\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2} and sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta, we can write:
sin⁡(π6)cos⁡θ−cos⁡(π6)sin⁡θ=sin⁡(2θ)\sin\left(\dfrac{\pi}{6}\right) \cos \theta - \cos\left(\dfrac{\pi}{6}\right) \sin \theta = \sin(2\theta)

⇒sin⁡(π6−θ)=sin⁡(2θ)\Rightarrow \sin\left(\dfrac{\pi}{6} - \theta\right) = \sin(2\theta)

Since θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right), we equate the principal angles:
π6−θ=2θ\dfrac{\pi}{6} - \theta = 2\theta

⇒3θ=π6\Rightarrow 3\theta = \dfrac{\pi}{6}

⇒θ=π18\Rightarrow \theta = \dfrac{\pi}{18}

Answer: π18\dfrac{\pi}{18}
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