Let $$f : (-1, 1) \rightarrow R$$ be a continuous function. If $$\int_0^{\sin x} f(t) dt = \frac{\sqrt{3}}{2}x$$, then $$f\left(\frac{\sqrt{3}}{2}\right)$$ is equal to:

We are given that $$ f : (-1, 1) \rightarrow \mathbb{R} $$ is a continuous function and satisfies the equation:

$$ \int_0^{\sin x} f(t) dt = \frac{\sqrt{3}}{2} x $$

We need to find $$ f\left( \frac{\sqrt{3}}{2} \right) $$.

Since $$ f $$ is continuous, we can use the Fundamental Theorem of Calculus. Let $$ F(x) = \int_0^x f(t) dt $$. Then the given equation becomes:

$$ F(\sin x) = \frac{\sqrt{3}}{2} x $$

Differentiate both sides with respect to $$ x $$. The derivative of the left side is found using the chain rule:

$$ \frac{d}{dx} F(\sin x) = F'(\sin x) \cdot \frac{d}{dx}(\sin x) = f(\sin x) \cdot \cos x $$

because $$ F'(u) = f(u) $$ by the Fundamental Theorem of Calculus.

The derivative of the right side is:

$$ \frac{d}{dx} \left( \frac{\sqrt{3}}{2} x \right) = \frac{\sqrt{3}}{2} $$

So we have:

$$ f(\sin x) \cos x = \frac{\sqrt{3}}{2} $$

Solving for $$ f(\sin x) $$:

$$ f(\sin x) = \frac{\frac{\sqrt{3}}{2}}{\cos x} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\cos x} = \frac{\sqrt{3}}{2} \sec x $$

Now, we want $$ f\left( \frac{\sqrt{3}}{2} \right) $$. Set $$ \sin x = \frac{\sqrt{3}}{2} $$. We know that $$ \sin x = \frac{\sqrt{3}}{2} $$ when $$ x = \frac{\pi}{3} $$ (considering the principal value in the interval where $$ x $$ is such that $$ \sin x $$ is in $$(-1,1)$$).

Substitute $$ x = \frac{\pi}{3} $$ into the expression:

$$ f\left( \sin \frac{\pi}{3} \right) = f\left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2} \sec \frac{\pi}{3} $$

We know that $$ \cos \frac{\pi}{3} = \frac{1}{2} $$, so $$ \sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2 $$.

Therefore:

$$ f\left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2} \cdot 2 = \sqrt{3} $$

Hence, the correct answer is Option B.