For $$a > 0$$, $$t \in \left(0, \frac{\pi}{2}\right)$$, let $$x = \sqrt{a^{\sin^{-1}t}}$$ and $$y = \sqrt{a^{\cos^{-1}t}}$$. Then, $$1 + \left(\frac{dy}{dx}\right)^2$$ equals :

Given $$ a > 0 $$ and $$ t \in \left(0, \frac{\pi}{2}\right) $$, we have $$ x = \sqrt{a^{\sin^{-1}t}} $$ and $$ y = \sqrt{a^{\cos^{-1}t}} $$. First, simplify the expressions for $$ x $$ and $$ y $$.

Rewrite $$ x $$ as: $$ x = \left( a^{\sin^{-1}t} \right)^{1/2} = a^{\frac{1}{2} \sin^{-1}t} $$ Similarly, rewrite $$ y $$ as: $$ y = \left( a^{\cos^{-1}t} \right)^{1/2} = a^{\frac{1}{2} \cos^{-1}t} $$

To find $$ \frac{dy}{dx} $$, use the chain rule since both $$ x $$ and $$ y $$ are functions of $$ t $$: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$

Compute $$ \frac{dx}{dt} $$ first. Set $$ u = \frac{1}{2} \sin^{-1}t $$, so $$ x = a^u $$. The derivative of $$ a^u $$ with respect to $$ u $$ is $$ a^u \ln a $$, and the derivative of $$ u $$ with respect to $$ t $$ is: $$ \frac{du}{dt} = \frac{1}{2} \cdot \frac{d}{dt} \left( \sin^{-1}t \right) = \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$ Thus: $$ \frac{dx}{dt} = \frac{dx}{du} \cdot \frac{du}{dt} = a^u \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$ Since $$ a^u = a^{\frac{1}{2} \sin^{-1}t} = x $$, substitute: $$ \frac{dx}{dt} = x \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} $$

Next, compute $$ \frac{dy}{dt} $$. Set $$ v = \frac{1}{2} \cos^{-1}t $$, so $$ y = a^v $$. The derivative of $$ a^v $$ with respect to $$ v $$ is $$ a^v \ln a $$, and the derivative of $$ v $$ with respect to $$ t $$ is: $$ \frac{dv}{dt} = \frac{1}{2} \cdot \frac{d}{dt} \left( \cos^{-1}t \right) = \frac{1}{2} \cdot \left( -\frac{1}{\sqrt{1-t^2}} \right) $$ Thus: $$ \frac{dy}{dt} = \frac{dy}{dv} \cdot \frac{dv}{dt} = a^v \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) $$ Since $$ a^v = a^{\frac{1}{2} \cos^{-1}t} = y $$, substitute: $$ \frac{dy}{dt} = y \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) $$

Now find $$ \frac{dy}{dx} $$: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{ y \ln a \cdot \left( -\frac{1}{2\sqrt{1-t^2}} \right) }{ x \ln a \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-t^2}} } $$ Simplify by canceling common factors. The $$ \ln a $$ terms cancel, and $$ \frac{1}{2} $$ and $$ \frac{1}{\sqrt{1-t^2}} $$ also cancel, leaving: $$ \frac{dy}{dx} = \frac{ y \cdot (-1) }{ x \cdot 1 } = -\frac{y}{x} $$

Now compute $$ \left( \frac{dy}{dx} \right)^2 $$: $$ \left( \frac{dy}{dx} \right)^2 = \left( -\frac{y}{x} \right)^2 = \frac{y^2}{x^2} $$

Then $$ 1 + \left( \frac{dy}{dx} \right)^2 $$ is: $$ 1 + \frac{y^2}{x^2} = \frac{x^2}{x^2} + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2} $$

Comparing with the options, this matches option D. Hence, the correct answer is Option D.