Answer:
Option B
Explanation:
We have,
$\sin^{-1}\frac{\sqrt{3}}{2}+\sin^{-1}\sqrt{\frac{2}{3}}$
$=\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^{2}}+\frac{\sqrt{3}}{2}\sqrt{1-\left(\sqrt{\frac{3}{2}}\right)^{2}}\right]$
As 0< $\frac{\sqrt{3}}{2}=0.866, \sqrt{\frac{2}{3}}=0.816 \leq$ 1 and
$\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\sqrt{\frac{2}{3}}\right)^{2}=1.4167>1$
$[ \because \sin^{-1} x+\sin^{-1}y=\pi-\sin^{-1}$
$x-\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}, if 0<x,y\leq1$ and $x^{2}+y^{2} >1$]
= $\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{1-\frac{2}{3}^{}}+\frac{\sqrt{2}}{3}\sqrt{1-{\frac{3}{4}}}\right]$
= $\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{\frac{3-2}{3}^{}}+\frac{\sqrt{2}}{3}\sqrt{{\frac{4-3}{4}}}\right]$
=$\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}{\frac{1}{\sqrt{3}}^{}}+\frac{\sqrt{2}}{\sqrt{3}}{\frac{1}{2}}\right]$
$=\pi-\sin^{-1}\left[\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\right]$
