If for the solution curve y = f(x) of the differential equation $$\frac{dy}{dx}+(\tan x)y = \frac{2+\sec x}{(1+2\sec x)^{2}}$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$$, then $$f\left(\frac{\pi}{4}\right)$$ is equal to :

The given differential equation is a first-order linear ODE:

$$ \frac{dy}{dx} + (\tan x)y = \frac{2 + \sec x}{(1 + 2\sec x)^2} $$

To solve it, we find the integrating factor: $$IF = e^{\int \tan x \, dx} = e^{-\ln|\cos x|} = \sec x$$.

Multiplying both sides of the differential equation by this integrating factor gives

$$ \frac{d}{dx}(y \sec x) = \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} $$

We simplify the right-hand side by expressing it in terms of $$\cos x$$:

$$ \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} = \frac{\frac{1}{\cos x}\left(2 + \frac{1}{\cos x}\right)}{\left(1 + \frac{2}{\cos x}\right)^2} = \frac{\frac{2\cos x + 1}{\cos^2 x}}{\frac{(\cos x + 2)^2}{\cos^2 x}} = \frac{2\cos x + 1}{(\cos x + 2)^2} $$

Observing that

$$ \frac{2\cos x + 1}{(\cos x + 2)^2} = \frac{d}{dx}\left(\frac{\sin x}{\cos x + 2}\right), $$

which can be checked by differentiating:

$$\frac{d}{dx}\left(\frac{\sin x}{\cos x + 2}\right) = \frac{\cos x(\cos x + 2) + \sin^2 x}{(\cos x + 2)^2} = \frac{\cos^2 x + 2\cos x + \sin^2 x}{(\cos x + 2)^2} = \frac{1 + 2\cos x}{(\cos x + 2)^2}.$$

Therefore, integrating both sides with respect to $$x$$ yields

$$ y\sec x = \frac{\sin x}{\cos x + 2} + C. $$

To determine the constant $$C$$, we use the condition $$f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$$:

$$ \frac{\sqrt{3}}{10} \cdot \sec\frac{\pi}{3} = \frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3} + 2} + C $$

$$ \frac{\sqrt{3}}{10} \cdot 2 = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2} + 2} + C $$

$$ \frac{2\sqrt{3}}{10} = \frac{\frac{\sqrt{3}}{2}}{\frac{5}{2}} + C = \frac{\sqrt{3}}{5} + C $$

$$ \frac{\sqrt{3}}{5} = \frac{\sqrt{3}}{5} + C $$

$$ C = 0 $$

Hence, $$y\sec x = \frac{\sin x}{\cos x + 2},$$ so that $$y = \frac{\sin x \cos x}{\cos x + 2}$$.

Evaluating this at $$x = \frac{\pi}{4}$$ gives

$$ f\left(\frac{\pi}{4}\right) = \frac{\sin\frac{\pi}{4} \cdot \cos\frac{\pi}{4}}{\cos\frac{\pi}{4} + 2} = \frac{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + 2} = \frac{\frac{1}{2}}{\frac{1+2\sqrt{2}}{\sqrt{2}}} = \frac{\sqrt{2}}{2(1+2\sqrt{2})} $$

Rationalizing the denominator gives

$$= \frac{\sqrt{2}(2\sqrt{2}-1)}{2(2\sqrt{2}+1)(2\sqrt{2}-1)} = \frac{\sqrt{2}(2\sqrt{2}-1)}{2(8-1)} = \frac{2\cdot 2 - \sqrt{2}}{14} = \frac{4-\sqrt{2}}{14} $$

The correct answer is Option 4: $$\frac{4-\sqrt{2}}{14}$$.