Let a vertical tower $$AB$$ of height $$2h$$ stands on a horizontal ground. Let from a point $$P$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2\alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\overrightarrow{AP}$$, he can see the top of the tower with an angle of elevation $$\alpha$$. If $$d = \sqrt{7}h$$, then $$\tan \alpha$$ is equal to

The differential equation is $$\frac{dy}{dx} + (\tan x)y = \sin x$$ with $$y(0) = 0$$. To solve it, we first compute the integrating factor: $$\text{IF} = e^{\int \tan x\, dx} = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x.$$

Multiplying the differential equation by this integrating factor gives $$\sec x \frac{dy}{dx} + \sec x \tan x \cdot y = \sec x \sin x = \tan x,$$ and the left-hand side can be recognized as the derivative of $$y\sec x$$, so $$\frac{d}{dx}(y \sec x) = \tan x.$$

Integrating both sides with respect to $$x$$ yields $$y \sec x = \int \tan x\, dx = -\ln|\cos x| + C = \ln|\sec x| + C.$$ Applying the initial condition $$y(0)=0$$ leads to $$0 \cdot \sec 0 = \ln|\sec 0| + C\quad\Longrightarrow\quad C = 0.$$ Hence $$y \sec x = \ln(\sec x)$$ and thus $$y = \cos x \cdot \ln(\sec x).$$

Finally, evaluating at $$x = \frac{\pi}{4}$$ gives $$y\left(\frac{\pi}{4}\right) = \cos\frac{\pi}{4} \cdot \ln\left(\sec\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \cdot \ln(\sqrt{2}) = \frac{1}{\sqrt{2}} \cdot \frac{1}{2}\ln 2 = \frac{\ln 2}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}\log_e 2.$$ Therefore, the answer is Option C: $$\dfrac{1}{2\sqrt{2}}\log_e 2$$.