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Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 2 April, Evening Shift — Question 35

Let P={θ∈[0,4π]:tan⁡2θ≠1}\mathrm{P} = \{\theta \in [0,4\pi ]:\tan^{2}\theta \neq 1\} and S={a∈Z:2(cos⁡8θ−sin⁡8θ)sec⁡2θ=a2,θ∈P}\mathrm{S} = \{\mathrm{a}\in \mathrm{Z}:2(\cos^{8}\theta -\sin^{8}\theta)\sec 2\theta = \mathrm{a}^{2},\theta \in \mathrm{P}\}. Then n(S)\mathrm{n}(\mathrm{S}) is:

  1. Option A:

    0

    Correct
  2. Option B:

    1

  3. Option C:

    2

  4. Option D:

    3

Answer: A

Step-by-step solution

2(cos⁡8θ−sin⁡8θ)sec⁡2θ=a22\left(\cos ^{8} \theta-\sin ^{8} \theta\right) \sec 2 \theta=\mathrm{a}^{2}

2(cos⁡4θ+sin⁡4θ)(cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)sec⁡2θ=a22(cos⁡4θ+sin⁡4θ)=a22(1−2sin⁡2θcos⁡2θ)=a22(1−sin⁡22θ2)=a2,a∈Za2=2−sin⁡22θ∈[1,2]a2=1 at sin⁡22θ=12θ=(2n+1)π2θ=(2n+1)π4∉Pn( S)=0\begin{aligned} & 2\left(\cos ^{4} \theta+\sin ^{4} \theta\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \sec 2 \theta=\mathrm{a}^{2} \\& 2\left(\cos ^{4} \theta+\sin ^{4} \theta\right)=\mathrm{a}^{2} \\& 2\left(1-2 \sin ^{2} \theta \cos ^{2} \theta\right)=\mathrm{a}^{2} \\& 2\left(1-\frac{\sin ^{2} 2 \theta}{2}\right)=\mathrm{a}^{2}, \mathrm{a} \in \mathrm{Z} \\& \mathrm{a}^{2}=2-\sin ^{2} 2 \theta \in[1,2] \\& \mathrm{a}^{2}=1 \text { at } \sin ^{2} 2 \theta=1 \\& 2 \theta=(2 \mathrm{n}+1) \frac{\pi}{2} \\& \theta=(2 \mathrm{n}+1) \frac{\pi}{4} \notin \mathrm{P} \\& \mathrm{n}(\mathrm{~S})=0 \end{aligned}

Answer key and solution verified before publishing.

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