JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 13

Simplify. $\cos^8\theta-\sin^8\theta=\cos2\theta\left(1-\frac12\sin^22\theta\right)$, so
$$2(\cos^8\theta-\sin^8\theta)\sec2\theta=2-\sin^22\theta.$$

Range. $\tan^2\theta\ne1$ means $\cos2\theta\ne0$, so $\sin^22\theta\in[0,1)$ and $a^2\in(1,2]$.

No integer $a$ has $a^2$ in $(1,2]$, so $n(S)=0$.

(A)