Find Number of Integer Elements in Set S Using Trigonometric Identity

To find the number of elements in the set $S$, we begin by simplifying the given trigonometric expression for $\theta \in P$:

$E(\theta) = 2(\cos^8\theta - \sin^8\theta) \sec 2\theta$

Step 1: Simplify the Trigonometric Expression

Using the algebraic identity for the difference of powers, $x^8 - y^8 = (x^4 - y^4)(x^4 + y^4) = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)$, we factorize $\cos^8\theta - \sin^8\theta$:

$\cos^8\theta - \sin^8\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)(\cos^4\theta + \sin^4\theta)$

Using the standard identities $\cos^2\theta + \sin^2\theta = 1$ and $\cos^2\theta - \sin^2\theta = \cos 2\theta$, this simplifies to:

$\cos^8\theta - \sin^8\theta = \cos 2\theta (\cos^4\theta + \sin^4\theta)$

Next, we express $\cos^4\theta + \sin^4\theta$ in terms of $\sin 2\theta$:

$\cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\cos^2\theta\sin^2\theta = 1 - \frac{1}{2}(2\sin\theta\cos\theta)^2 = 1 - \frac{1}{2}\sin^2 2\theta$

Substituting this back into our expression:

$\cos^8\theta - \sin^8\theta = \cos 2\theta \left(1 - \frac{1}{2}\sin^2 2\theta\right)$

Now, substituting this back into $E(\theta)$:

$E(\theta) = 2 \cdot \cos 2\theta \left(1 - \frac{1}{2}\sin^2 2\theta\right) \sec 2\theta$

Since $\theta \in P$, we have $\tan^2\theta \neq 1$, which implies $\cos 2\theta \neq 0$. Therefore, $\cos 2\theta \cdot \sec 2\theta = 1$. Thus:

$E(\theta) = 2\left(1 - \frac{1}{2}\sin^2 2\theta\right) = 2 - \sin^2 2\theta$

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Step 2: Find the Range of $E(\theta)$

Since $\sin^2 2\theta \ge 0$ for all $\theta$, and for $\theta \in P$, $\tan^2\theta \neq 1 \implies 2\theta \neq (2k+1)\frac{\pi}{2} \implies \sin^2 2\theta \neq 1$.

Therefore, the range of $\sin^2 2\theta$ for $\theta \in P$ is:

$0 \le \sin^2 2\theta < 1$

Subtracting this from $2$ gives:

$1 < 2 - \sin^2 2\theta \le 2$

Thus, the range of $E(\theta)$ is:

$E(\theta) \in (1, 2]$

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Step 3: Determine $n(S)$

The set $S$ is defined as $S = \{a \in \mathbb{Z} : a^2 = E(\theta), \theta \in P\}$.

From the range of $E(\theta)$, we must have:

$a^2 \in (1, 2]$

Since $a \in \mathbb{Z}$, $a^2$ must be a perfect square integer (i.e., $a^2 \in \{0, 1, 4, 9, 16, \dots\}$).

None of these perfect square integers lie in the interval $(1, 2]$. Consequently, there is no integer $a$ that satisfies the given condition.

Thus, $S = \emptyset$, which means:

$n(S) = 0$

Correct Option: A