If the solution curve of the differential equation $$((\tan^{-1}y) - x)dy = (1 + y^2)dx$$ passes through the point $$(1, 0)$$ then the abscissa of the point on the curve whose ordinate is $$\tan(1)$$ is

We need to solve the differential equation $$(\tan^{-1}y - x)\,dy = (1 + y^2)\,dx$$.

Rewriting: $$\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1 + y^2}$$

$$\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$$

This is a linear ODE in $$x$$ with $$P(y) = \frac{1}{1+y^2}$$ and $$Q(y) = \frac{\tan^{-1}y}{1+y^2}$$.

Integrating factor: $$\mu = e^{\int \frac{dy}{1+y^2}} = e^{\tan^{-1}y}$$

The solution is:

$$x \cdot e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} \cdot e^{\tan^{-1}y}\,dy$$

Let $$t = \tan^{-1}y$$, so $$dt = \frac{dy}{1+y^2}$$:

$$x \cdot e^{\tan^{-1}y} = \int t \cdot e^t \, dt = t \cdot e^t - e^t + C = e^t(t - 1) + C$$

$$x \cdot e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C$$

Using the initial condition $$(x, y) = (1, 0)$$: $$\tan^{-1}(0) = 0$$, so:

$$1 \cdot e^0 = e^0(0 - 1) + C \implies 1 = -1 + C \implies C = 2$$

Therefore: $$x \cdot e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + 2$$

$$x = (\tan^{-1}y - 1) + 2e^{-\tan^{-1}y}$$

Now we find $$x$$ when $$y = \tan(1)$$, so $$\tan^{-1}y = 1$$:

$$x = (1 - 1) + 2e^{-1} = \frac{2}{e}$$

The correct answer is Option B: $$\frac{2}{e}$$.