JEE Main 2026 — Trigonometric Ratios & Identities Question with Solution

Given the expression $2({\cos }^8\theta -{\sin }^8\theta )\sec 2\theta =a^2$.

We can simplify the term ${\cos }^8\theta -{\sin }^8\theta$ as follows:
${\cos }^8\theta -{\sin }^8\theta =({\cos }^4\theta -{\sin }^4\theta )({\cos }^4\theta +{\sin }^4\theta )$
$\Rightarrow ({\cos }^2\theta -{\sin }^2\theta )({\cos }^2\theta +{\sin }^2\theta )(({\cos }^2\theta +{\sin }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta )$
$\Rightarrow \cos 2\theta \times 1\times \left(1-\frac{1}{2}{\sin }^{2}2\theta \right)$

Substitute this back into the given equation:
$2\left[\cos 2\theta \left(1-\frac{1}{2}{\sin }^{2}2\theta \right)\right]\frac{1}{\cos 2\theta }=a^2$

Since $\theta \in P$, we have ${\tan }^2\theta \ne 1$, which implies $\cos 2\theta \ne 0$. Thus, we can safely cancel $\cos 2\theta$:
$2\left(1-\frac{1}{2}{\sin }^{2}2\theta \right)=a^2$
$\Rightarrow 2-{\sin }^{2}2\theta =a^2$

We know that for any real $\theta$, $0\le {\sin }^{2}2\theta \le 1$.
However, since ${\tan }^2\theta \ne 1$, ${\sin }^{2}2\theta \ne 1$.
Therefore, $0\le {\sin }^{2}2\theta <1$.

Multiplying by $-1$ and adding $2$, we get the range of $a^2$:
$1<2-{\sin }^{2}2\theta \le 2$
$\Rightarrow 1<a^2\le 2$

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

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

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

Therefore, $n(S)=0$.

Answer: $0$