Solution of Triangles & Trig EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Trigonometric Identity Set Problem: n(S) = 0 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let P={θ∈[0,4π]:tan⁡2θ≠1}P=\{\theta\in[0,4\pi]:\tan^2\theta\ne 1\} and S={a∈Z:2(cos⁡8θ−sin⁡8θ)sec⁡2θ=a2, θ∈P}S=\{a\in\mathbb{Z}:2(\cos^8\theta-\sin^8\theta)\sec2\theta=a^2,\ \theta\in P\}. Then n(S)n(S) is
A00correct
B11
C22
D33
Solution
Step 1: Factor cos⁡8θ−sin⁡8θ\cos^8\theta-\sin^8\theta:
cos⁡8θ−sin⁡8θ=(cos⁡4θ+sin⁡4θ)(cos⁡4θ−sin⁡4θ)=(cos⁡4θ+sin⁡4θ)(cos⁡2θ−sin⁡2θ).\cos^8\theta-\sin^8\theta=(\cos^4\theta+\sin^4\theta)(\cos^4\theta-\sin^4\theta)=(\cos^4\theta+\sin^4\theta)(\cos^2\theta-\sin^2\theta).
 Since cos⁡2θ−sin⁡2θ=cos⁡2θ\cos^2\theta-\sin^2\theta=\cos2\theta,
2(cos⁡8θ−sin⁡8θ)sec⁡2θ=2(cos⁡4θ+sin⁡4θ)cos⁡2θ⋅1cos⁡2θ=2(cos⁡4θ+sin⁡4θ).2(\cos^8\theta-\sin^8\theta)\sec2\theta=2(\cos^4\theta+\sin^4\theta)\cos2\theta\cdot\frac{1}{\cos2\theta}=2(\cos^4\theta+\sin^4\theta).
 Step 2: Use cos⁡4θ+sin⁡4θ=1−2sin⁡2θcos⁡2θ=1−sin⁡22θ2\cos^4\theta+\sin^4\theta=1-2\sin^2\theta\cos^2\theta=1-\dfrac{\sin^2 2\theta}{2}:
2(1−sin⁡22θ2)=2−sin⁡22θ=a2.2\left(1-\frac{\sin^2 2\theta}{2}\right)=2-\sin^2 2\theta=a^2.
 Step 3: Since sin⁡22θ∈[0,1]\sin^2 2\theta\in[0,1], we have a2=2−sin⁡22θ∈[1,2]a^2=2-\sin^2 2\theta\in[1,2]. The only integer value of a2a^2 in [1,2][1,2] is a2=1a^2=1, which needs sin⁡22θ=1\sin^2 2\theta=1. Step 4: sin⁡22θ=1⇒2θ=(2n+1)π2⇒θ=(2n+1)π4\sin^2 2\theta=1\Rightarrow 2\theta=(2n+1)\dfrac{\pi}{2}\Rightarrow\theta=(2n+1)\dfrac{\pi}{4}. But at these θ\theta, tan⁡2θ=1\tan^2\theta=1, so they are excluded from PP. Step 5: No admissible θ\theta remains, hence
n(S)=0.n(S)=0.
 Correct answer: (1)
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