If $$f(x) = \sin(\sin x)$$ and $$f''(x) + \tan x \cdot f'(x) + g(x) = 0$$, then $$g(x)$$ is :

We are given the function $$ f(x) = \sin(\sin x) $$ and the equation $$ f''(x) + \tan x \cdot f'(x) + g(x) = 0 $$. We need to find $$ g(x) $$.

First, we find the first derivative of $$ f(x) $$. Using the chain rule, let $$ u = \sin x $$, so $$ f(x) = \sin u $$. Then:

$$ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \cos u \cdot \cos x $$

Substituting $$ u = \sin x $$ back in:

$$ f'(x) = \cos(\sin x) \cdot \cos x $$

Next, we find the second derivative $$ f''(x) $$. Since $$ f'(x) = \cos(\sin x) \cdot \cos x $$ is a product of two functions, we use the product rule:

$$ f''(x) = \frac{d}{dx} \left[ \cos(\sin x) \right] \cdot \cos x + \cos(\sin x) \cdot \frac{d}{dx} \left[ \cos x \right] $$

Compute each derivative separately. For $$ \frac{d}{dx} \left[ \cos(\sin x) \right] $$, let $$ v = \sin x $$, so:

$$ \frac{d}{dx} \cos v = -\sin v \cdot \frac{dv}{dx} = -\sin(\sin x) \cdot \cos x $$

And $$ \frac{d}{dx} \left[ \cos x \right] = -\sin x $$. Substituting these back:

$$ f''(x) = \left[ -\sin(\sin x) \cdot \cos x \right] \cdot \cos x + \cos(\sin x) \cdot \left[ -\sin x \right] $$

Simplify:

$$ f''(x) = -\sin(\sin x) \cdot \cos^2 x - \cos(\sin x) \cdot \sin x $$

Now, substitute $$ f''(x) $$ and $$ f'(x) $$ into the given equation:

$$ \left[ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x \right] + \tan x \cdot \left[ \cos(\sin x) \cos x \right] + g(x) = 0 $$

We know $$ \tan x = \frac{\sin x}{\cos x} $$, so simplify the term with $$ \tan x $$:

$$ \tan x \cdot f'(x) = \frac{\sin x}{\cos x} \cdot \cos(\sin x) \cos x = \sin x \cdot \cos(\sin x) $$

Substitute this back into the equation:

$$ -\sin(\sin x) \cos^2 x - \cos(\sin x) \sin x + \sin x \cos(\sin x) + g(x) = 0 $$

Notice that $$ - \cos(\sin x) \sin x + \sin x \cos(\sin x) = 0 $$, so they cancel out:

$$ -\sin(\sin x) \cos^2 x + g(x) = 0 $$

Therefore, solving for $$ g(x) $$:

$$ g(x) = \sin(\sin x) \cos^2 x $$

We can write this as:

$$ g(x) = \cos^2 x \cdot \sin(\sin x) $$

Comparing with the options:

A. $$ \cos^2 x \cos(\sin x) $$

B. $$ \sin^2 x \cos(\cos x) $$

C. $$ \sin^2 x \sin(\cos x) $$

D. $$ \cos^2 x \sin(\sin x) $$

Our expression matches option D.

Hence, the correct answer is Option D.