Question Stimulus :-
If `sin^(-1)`1/3 + `sin^(-1)`2/3=`sin^(-1)x`, then x=?

Options

A. 0
B. `(sqrt5-4sqrt2)/9`
C. `(sqrt5+4sqrt2)/9`
D. `pi/2`

Answer

To solve the equation \( \sin^{-1} \left( \frac{1}{3} \right) + \sin^{-1} \left( \frac{2}{3} \right) = \sin^{-1}(x) \), we will use the formula for the sum of inverse sine functions:

\[ \sin^{-1}(a) + \sin^{-1}(b) = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right) \]

### Step 1: Identify \( a \) and \( b \)

In our case, we have:
- \( a = \frac{1}{3} \)
- \( b = \frac{2}{3} \)

### Step 2: Calculate \( \sqrt{1 - b^2} \) and \( \sqrt{1 - a^2} \)

Now, we need to calculate:
- \( \sqrt{1 - b^2} = \sqrt{1 - \left( \frac{2}{3} \right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \)

- \( \sqrt{1 - a^2} = \sqrt{1 - \left( \frac{1}{3} \right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \)

### Step 3: Substitute into the formula

Now, substitute these values into the formula:

\[ \sin^{-1} \left( \frac{1}{3} \right) + \sin^{-1} \left( \frac{2}{3} \right) = \sin^{-1} \left( \frac{1}{3} \cdot \frac{\sqrt{5}}{3} + \frac{2}{3} \cdot \frac{2\sqrt{2}}{3} \right) \]

### Step 4: Simplify the expression

Calculating the expression inside the sine inverse:

\[ = \sin^{-1} \left( \frac{\sqrt{5}}{9} + \frac{4\sqrt{2}}{9} \right) \]

Combining the terms:

\[ = \sin^{-1} \left( \frac{\sqrt{5} + 4\sqrt{2}}{9} \right) \]

### Step 5: Set equal to \( \sin^{-1}(x) \)

Since we have \( \sin^{-1}(x) = \sin^{-1} \left( \frac{\sqrt{5} + 4\sqrt{2}}{9} \right) \), we can equate the arguments:

\[ x = \frac{\sqrt{5} + 4\sqrt{2}}{9} \]

### Final Answer

Thus, the value of \( x \) is:

\[ \boxed{\frac{\sqrt{5} + 4\sqrt{2}}{9}} \]