Let Z denote the set of integers. Then match | vedclass

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(C) We have $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$.Since $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\cos ^{-1}\left(\f

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$A-V, B-I, C-IV, D-II$

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(C) We have $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}ight)+\sin ^{-1} \frac{1}{3}$.Since $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}ight)=\cos ^{-1}\left(\frac{1}{3}ight)$, we have $\cos ^{-1}\left(\frac{1}{3}ight)+\sin ^{-1}\left(\frac{1}{3}ight)=\frac{\pi}{2}$. Thus, $A-V$.$(B)$ Let $	heta=\sin ^{-1}\left(\frac{(-1)^n}{2}ight), n \in Z$. Then $\sin 	heta=\frac{(-1)^n}{2}=\sin\left((-1)^n \frac{\pi}{6}ight)$. The general solution is $	heta=k \pi+(-1)^k\left((-1)^n \frac{\pi}{6}ight)$, which simplifies to $k \pi \pm \frac{\pi}{6}, k \in Z$. This matches $I$.$(C)$ Let $\alpha=	an ^{-1}\left(\sec \frac{\pi}{4}+	an \frac{\pi}{4}ight) = 	an ^{-1}(\sqrt{2}+1)$. Since $	an \frac{3\pi}{8} = \sqrt{2}+1$, we have $\alpha = \frac{3\pi}{8}$. Thus, $C-IV$.$(D)$ Let $t=\sin ^{-1}|\sin x|$. Then $t=\sqrt{t} \Rightarrow t^2-t=0 \Rightarrow t(t-1)=0$. So $t=0$ or $t=1$. $\sin ^{-1}|\sin x|=0 \Rightarrow |\sin x|=0 \Rightarrow x=k\pi$. $\sin ^{-1}|\sin x|=1 \Rightarrow |\sin x|=\sin 1 \Rightarrow x=k\pi \pm 1$. Combining these, the general solution is $x=k\pi \pm 1, k \in Z$. Thus, $D-II$.

List-$I$	List-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

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