Let S = { theta in [0, 4pi] : tan^2 theta neq | vedclass

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(C) Given the expression $2(\cos^8 	heta - \sin^8 	heta) \sec 2	heta = a^2$.Using the difference of squares,$\cos^8 	heta - \sin^8 	heta = (\cos^

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$2$

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(C) Given the expression $2(\cos^8 	heta - \sin^8 	heta) \sec 2	heta = a^2$.Using the difference of squares,$\cos^8 	heta - \sin^8 	heta = (\cos^4 	heta - \sin^4 	heta)(\cos^4 	heta + \sin^4 	heta) = (\cos^2 	heta - \sin^2 	heta)(\cos^2 	heta + \sin^2 	heta)(\cos^4 	heta + \sin^4 	heta) = \cos 2	heta (\cos^4 	heta + \sin^4 	heta)$.Substituting this into the expression: $2 \cos 2	heta (\cos^4 	heta + \sin^4 	heta) \cdot \frac{1}{\cos 2	heta} = 2(\cos^4 	heta + \sin^4 	heta) = a^2$.We know that $\cos^4 	heta + \sin^4 	heta = (\cos^2 	heta + \sin^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - \frac{1}{2} \sin^2 2	heta$.Thus,$a^2 = 2(1 - \frac{1}{2} \sin^2 2	heta) = 2 - \sin^2 2	heta$.Since $0 \le \sin^2 2	heta \le 1$,we have $2 - 1 \le a^2 \le 2 - 0$,which means $1 \le a^2 \le 2$.Since $a \in \mathbb{Z}$,$a^2$ must be a perfect square. The only perfect square in the range $[1, 2]$ is $1$.Therefore,$a^2 = 1$,which implies $a = 1$ or $a = -1$.Thus,the set $S$ contains $2$ elements.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

Let $S = \{ \theta \in [0, 4\pi] : \tan^2 \theta \neq 1 \}$ and $S = \{ a \in \mathbb{Z} : 2(\cos^8 \theta - \sin^8 \theta) \sec 2\theta = a^2, \theta \in S \}$. Then $n(S)$ is:

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