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Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 6 April, Morning Shift — Question 34

Let S={θ∈(−2π,2π):cos⁡θ+1=3sin⁡θ}\mathrm{S}=\{\theta \in(-2 \pi, 2 \pi): \cos \theta+1=\sqrt{3} \sin \theta\}. Then ∑θ∈Sθ\sum_{\theta \in \mathrm{S}} \theta is equal to:

  1. Option A:

    −2π3-\frac{2 \pi}{3}

  2. Option B:

    −4π3-\frac{4 \pi}{3}

    Correct
  3. Option C:

    2π3\frac{2 \pi}{3}

  4. Option D:

    4π3\frac{4 \pi}{3}

Answer: B

Step-by-step solution

cos⁡θ+1=3sin⁡θ\cos \theta+1=\sqrt{3} \sin \theta 1−tan⁡2θ21+tan⁡2θ2+1=3(2tan⁡θ21+tan⁡2θ2)\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}+1=\sqrt{3}\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right) 2=23tan⁡θ22=2 \sqrt{3} \tan \frac{\theta}{2} ⇒tan⁡θ2=13;θ∈(−2π,2π),θ2∈(−π,π)\Rightarrow \tan \frac{\theta}{2}=\frac{1}{\sqrt{3}} ;\theta \in(-2 \pi, 2 \pi), \frac{\theta}{2} \in(-\pi, \pi) θ2=−5π6,π6\frac{\theta}{2}=-\frac{5 \pi}{6}, \frac{\pi}{6} θ=−5π3,π3\theta=\frac{-5 \pi}{3}, \frac{\pi}{3} Sum =−−5π3+π3=−4π3=-\frac{-5 \pi}{3}+\frac{\pi}{3}=\frac{-4 \pi}{3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Periodicity of trigometric functions,Solutions of trigonometric equations