Let P = {θ ∈ [0, 4π] : tan²θ ≠ 1} and S = {a ∈ ℤ : 2(cos⁸θ − sin⁸θ)sec2θ = a², θ ∈ P}. Then n(S) is:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2026

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Home Maths Trigonometric FunctionsQuestion #495

Let

P = {θ \in [0, 4 π] : \tan^{2} θ \neq 1}

and

S = {a \in Z : 2 (\cos^{8} θ - \sin^{8} θ) \sec 2 θ = a^{2}, θ \in P}

. Then

n (S)

is:

Clue 1 of 3: Problem Structure

The problem requires simplifying the complex trigonometric expression to find its range, and then checking which integer squares fall within that range, considering the given domain constraints.

🥷

Ninja Strategy•Quick Range Check

The equation $a^{2} = 2 - \sin^{2} (2 θ)$ implies $1 \leq a^{2} \leq 2$ . The only integer square in this range is 1. A quick check shows $a^{2} = 1$ is disallowed by the domain, so no solutions exist.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Simplify the Trigonometric Equation✦ Active

The given equation is $2 (\cos^{8} θ - \sin^{8} θ) \sec 2 θ = a^{2}$ . We simplify the term $\cos^{8} θ - \sin^{8} θ$ using the difference of squares formula.

\cos^{8} θ - \sin^{8} θ = (\cos^{4} θ - \sin^{4} θ) (\cos^{4} θ + \sin^{4} θ)

= (\cos^{2} θ - \sin^{2} θ) (\cos^{2} θ + \sin^{2} θ) ((\cos^{2} θ + \sin^{2} θ)^{2} - 2 \sin^{2} θ \cos^{2} θ)

= (\cos 2 θ) (1) (1 - 2 {(\frac{\sin 2 θ}{2})}^{2}) = \cos 2 θ (1 - \frac{\sin^{2} 2 θ}{2})

Substituting this back into the original equation and using $\sec 2 θ = 1 / \cos 2 θ$ :

a^{2} = 2 [\cos 2 θ (1 - \frac{\sin^{2} 2 θ}{2})] \frac{1}{\cos 2 θ} = 2 (1 - \frac{\sin^{2} 2 θ}{2}) = 2 - \sin^{2} 2 θ

Step 2: Determine the Range of

a^{2}

○ Expand

The domain for $θ$ is given by set P, where $θ \in [0, 4 π]$ and $\tan^{2} θ \neq 1$ . The condition $\tan^{2} θ \neq 1$ implies $\cos^{2} θ \neq \sin^{2} θ$ , which means $\cos^{2} θ - \sin^{2} θ \neq 0$ , so $\cos 2 θ \neq 0$ . This is consistent with the presence of $\sec 2 θ$ in the equation.

Since $\cos 2 θ \neq 0$ , we have $\cos^{2} 2 θ \neq 0$ , which means $1 - \sin^{2} 2 θ \neq 0$ , so $\sin^{2} 2 θ \neq 1$ .

The general range of $\sin^{2} (2 θ)$ is $[0, 1]$ . Due to the domain restriction, the range for $θ \in P$ becomes $[0, 1)$ . Now we find the range of $a^{2} = 2 - \sin^{2} 2 θ$ . Since $0 \leq \sin^{2} 2 θ < 1$ , we have $1 < 2 - \sin^{2} 2 θ \leq 2$ . Therefore, the range of $a^{2}$ is $(1, 2]$ .

💡 Teacher's Secret Hint

The strict inequality from the domain condition is crucial. Forgetting it would lead to an incorrect range for $a^{2}$ .

Step 3: Find Integer Solutions for

a

○ Expand

We are looking for integer values of $a$ (i.e., $a \in Z$ ) such that $a^{2}$ is in the interval $(1, 2]$ . We check the squares of integers: $0^{2} = 0$ , $(\pm 1)^{2} = 1$ , $(\pm 2)^{2} = 4$ , etc.

None of these perfect squares lie in the interval $(1, 2]$ . The value $1$ is not included in the interval, and there are no other perfect squares up to $2$ .

Thus, there are no integer values for $a$ that satisfy the condition. The set S is empty, so $n (S) = 0$ .

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Problem Structure

🧠 Full Solution Path