Question

If ${\sin }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{2}{3}={\sin }^{-1}x$, then the value of $x$ is-

• A. $0$
• B. $\frac{\sqrt{5}-4\sqrt{2}}{9}$
• C. $\frac{\sqrt{5}+2\sqrt{3}}{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{\sqrt{5}+2\sqrt{3}}{6}$

Given, ${\sin }^{-1}\frac{1}{2}+{\sin }^{-1}\frac{2}{3}={\sin }^{-1}x$
$\Rightarrow {\sin }^{-1}(\frac{1}{2}\sqrt{1-{\left(\frac{2}{3}\right)}^2}+\frac{2}{3}\sqrt{1-{\left(\frac{1}{2}\right)}^2})={\sin }^{-1}x$
$\Rightarrow {\sin }^{-1}(\frac{1}{2}\times \frac{\sqrt{5}}{3}+\frac{2}{3}\times \frac{\sqrt{3}}{2})={\sin }^{-1}x$
$\Rightarrow {\sin }^{-1}(\frac{\sqrt{5}}{6}+\frac{2\sqrt{3}}{6})={\sin }^{-1}x$
$\Rightarrow {\sin }^{-1}\left(\frac{\sqrt{5}+2\sqrt{3}}{6}\right)={\sin }^{-1}x\Rightarrow x=\frac{\sqrt{5}+2\sqrt{3}}{6}$