Question

If ${\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{2}{3}={\sin }^{-1}x$, then $x$ is equal to

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

Answer 1

The correct option is C

$\frac{(\sqrt{5}+4\sqrt{2})}{9}$

Explanation for the correct option.

Given that, ${\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{2}{3}={\sin }^{-1}x$

Now using the identity, $\left.\left.{\sin }^{-1}a+{\sin }^{-1}b={\sin }^{-1}\left[a\sqrt{(}1-b^2\right)+b\sqrt{(}1-a^2\right)\right]$.

Here, $a=\frac{1}{3},b=\frac{2}{3}$

Substituting values we get:
$\begin{array}{rcl}{\sin }^{-1}\frac{1}{3}+{\sin }^{-1}\frac{2}{3}& =& {\sin }^{-1}\left[\frac{1}{3}\sqrt{\left(1-\frac{4}{9}\right)}+\frac{2}{3}\sqrt{\left(1-\frac{1}{9}\right)}\right]\\ & =& {\sin }^{-1}\left[\frac{1}{3}\sqrt{\frac{5}{9}}+\frac{2}{3}\sqrt{\frac{8}{9}}\right]\\ & =& {\sin }^{-1}\left[\frac{1}{9}\sqrt{5}+\frac{4}{9}\sqrt{2}\right]\\ & =& {\sin }^{-1}\left[\frac{\sqrt{5}+4\sqrt{2}}{9}\right]\end{array}$

Hence, option C is correct.