Let $$x \in (0, 1)$$. The set of all $$x$$ such that $$\sin^{-1}x > \cos^{-1}x$$, is the interval:

We are given that $$ x \in (0, 1) $$ and need to find the set of all $$ x $$ such that $$ \sin^{-1} x > \cos^{-1} x $$.

Recall the identity: for any $$ x \in [-1, 1] $$, $$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $$. Therefore, we can express $$ \cos^{-1} x $$ as $$ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x $$.

Substitute this into the inequality:

$$ \sin^{-1} x > \cos^{-1} x $$

$$ \sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x $$

Add $$ \sin^{-1} x $$ to both sides:

$$ \sin^{-1} x + \sin^{-1} x > \frac{\pi}{2} $$

$$ 2 \sin^{-1} x > \frac{\pi}{2} $$

Divide both sides by 2:

$$ \sin^{-1} x > \frac{\pi}{4} $$

Since $$ x \in (0, 1) $$, $$ \sin^{-1} x $$ lies in $$ (0, \frac{\pi}{2}) $$, where the sine function is strictly increasing. Therefore, taking sine on both sides preserves the inequality:

$$ \sin(\sin^{-1} x) > \sin\left(\frac{\pi}{4}\right) $$

$$ x > \sin\left(\frac{\pi}{4}\right) $$

$$ x > \frac{1}{\sqrt{2}} $$

Also, since $$ x \in (0, 1) $$, we have $$ x < 1 $$. Thus, $$ x $$ must satisfy $$ \frac{1}{\sqrt{2}} < x < 1 $$, which is the interval $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$.

Now, verify the boundaries:

• At $$ x = \frac{1}{\sqrt{2}} $$, $$ \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} $$ and $$ \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} $$, so equality holds, not inequality.
• As $$ x $$ approaches 1 from the left, $$ \sin^{-1} x $$ approaches $$ \frac{\pi}{2} $$ and $$ \cos^{-1} x $$ approaches 0, so $$ \sin^{-1} x > \cos^{-1} x $$ holds.

Comparing with the options:

• A: $$ \left( \frac{1}{2}, \frac{1}{\sqrt{2}} \right) $$ → This includes values less than $$ \frac{1}{\sqrt{2}} $$, which do not satisfy the inequality.
• B: $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$ → Matches our solution.
• C: $$ (0, 1) $$ → Includes values less than or equal to $$ \frac{1}{\sqrt{2}} $$, which do not satisfy.
• D: $$ \left( 0, \frac{\sqrt{3}}{2} \right) $$ → $$ \frac{\sqrt{3}}{2} \approx 0.866 $$ and $$ \frac{1}{\sqrt{2}} \approx 0.707 $$, so this interval includes values below $$ \frac{1}{\sqrt{2}} $$ and excludes values above $$ \frac{\sqrt{3}}{2} $$ that are in $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$, hence incorrect.

Hence, the correct answer is Option B.