JEE Main 2026 — Mathematics Trigonometry

Given the expression $2(\cos^8\theta - \sin^8\theta)\sec 2\theta = a^2$2(cos8θ−sin8θ)sec2θ=a2.

We can simplify the term $\cos^8\theta - \sin^8\theta$cos8θ−sin8θ as follows:

$$\cos^8\theta - \sin^8\theta = (\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta)$$cos8θ−sin8θ=(cos4θ−sin4θ)(cos4θ+sin4θ)

$$\Rightarrow (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)((\cos^2\theta + \sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta)$$⇒(cos2θ−sin2θ)(cos2θ+sin2θ)((cos2θ+sin2θ)2−2sin2θcos2θ)

$$\Rightarrow \cos 2\theta \times 1 \times \left(1 - \dfrac{1}{2}\sin^2 2\theta\right)$$⇒cos2θ×1×(1−21​sin22θ)

Substitute this back into the given equation:

$$2 \left[ \cos 2\theta \left(1 - \dfrac{1}{2}\sin^2 2\theta\right) \right] \dfrac{1}{\cos 2\theta} = a^2$$2[cos2θ(1−21​sin22θ)]cos2θ1​=a2

Since $\theta \in P$θ∈P, we have $\tan^2\theta \neq 1$tan2θ=1, which implies $\cos 2\theta \neq 0$cos2θ=0. Thus, we can safely cancel $\cos 2\theta$cos2θ:

$$2 \left(1 - \dfrac{1}{2}\sin^2 2\theta\right) = a^2$$2(1−21​sin22θ)=a2

$$\Rightarrow 2 - \sin^2 2\theta = a^2$$⇒2−sin22θ=a2

We know that for any real $\theta$θ, $0 \le \sin^2 2\theta \le 1$0≤sin22θ≤1.

However, since $\tan^2\theta \neq 1$tan2θ=1, $\sin^2 2\theta \neq 1$sin22θ=1.

Therefore, $0 \le \sin^2 2\theta < 1$0≤sin22θ<1.

Multiplying by $-1$−1 and adding $2$2, we get the range of $a^2$a2:

$$1 < 2 - \sin^2 2\theta \le 2$$1<2−sin22θ≤2

$$\Rightarrow 1 < a^2 \le 2$$⇒1<a2≤2

Since $a \in \mathbb{Z}$a∈Z, $a^2$a2 must be a perfect square integer. The only integer in the interval $(1, 2]$(1,2] is $2$2, so $a^2 = 2$a2=2.

But there is no integer $a$a such that $a^2 = 2$a2=2.

Thus, the set $S$S contains no elements, meaning $S = \emptyset$S=∅.

Therefore, $n(S) = 0$n(S)=0.

Answer: $0$0