Given, $A=\{1,2,3,4,5,6\}$ $R=\{(x, y): y \;\text{ is divisible by }\; x\}$(i) $(2,4) \in R$    $\{ \because 4$ is divisible by 2 $\}$ But $(4,2)$∉ R $\{ \because 2$ is not divisible by 4$\}$ $\therefore \;\; R$ is not symmetric.   (ii) Let $(a, b) \in R \& (b, c) \in R$ $\Rightarrow \;\; b=\lambda a$ and $c=\mu b$Now, $c=\mu b=\mu(\lambda a) \Rightarrow (a, c) \in R$ $\Rightarrow \;\; c$ is divisible by $a$ $\therefore \;\; bo R$ is transitive.  OR $\;\text{ L.H.S. }\;=\;\frac{9\pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3}$ $=\;\frac{9}{4} \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{3} \right)$ $=\;\frac{9}{4} \left( \cos^{-1} \frac{1}{3} \right) \ldots (1)$Now, let $\cos^{-1} \frac{1}{3} = x$ .Then, $\cos x = \frac{1}{3} \Rightarrow \sin x = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \frac{2\sqrt{2}}{3}$ $\therefore x = \sin^{-1} \frac{2\sqrt{2}}{3} \Rightarrow \cos^{-1} \frac{1}{3} = \sin^{-1} \frac{2\sqrt{2}}{3}$ $\therefore \;\text{ L.H.S. }\;=\;\frac{9}{4} \sin^{-1} \frac{2\sqrt{2}}{3} = \;\text{ R.H.S. }\;$