If sin^{-1} frac{1}{3} + sin^{-1} frac{2}{3} | vedclass

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(C) We use the formula $\sin^{-1} a + \sin^{-1} b = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right)$.Given $\sin^{-1} \frac{1}{3} + \sin^

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

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(C) We use the formula $\sin^{-1} a + \sin^{-1} b = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right)$.Given $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2}{3} = \sin^{-1} x$.Applying the formula with $a = \frac{1}{3}$ and $b = \frac{2}{3}$:$\sin^{-1} x = \sin^{-1} \left( \frac{1}{3} \sqrt{1 - (\frac{2}{3})^2} + \frac{2}{3} \sqrt{1 - (\frac{1}{3})^2} \right)$$= \sin^{-1} \left( \frac{1}{3} \sqrt{1 - \frac{4}{9}} + \frac{2}{3} \sqrt{1 - \frac{1}{9}} \right)$$= \sin^{-1} \left( \frac{1}{3} \sqrt{\frac{5}{9}} + \frac{2}{3} \sqrt{\frac{8}{9}} \right)$$= \sin^{-1} \left( \frac{1}{3} \cdot \frac{\sqrt{5}}{3} + \frac{2}{3} \cdot \frac{2\sqrt{2}}{3} \right)$$= \sin^{-1} \left( \frac{\sqrt{5} + 4\sqrt{2}}{9} \right)$.Therefore,$x = \frac{\sqrt{5} + 4\sqrt{2}}{9}$.

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

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