Given differential equation

$\big((\tan^{-1} y)-x\big)\,dy=(1+y^2)\,dx$

We want the solution curve passing through $(1,0)$.

Rewrite in a more convenient form

Divide both sides by $dy$:

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

So,

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

This is a linear differential equation in $x$ as a function of $y$.

Find the integrating factor

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

Multiply throughout by the integrating factor

Integrate both sides

Let

$t=\tan^{-1}y \implies dt=\frac{1}{1+y^2}dy$

Then,

$\int \frac{(\tan^{-1}y)e^{\tan^{-1}y}}{1+y^2}\,dy=\int te^t\,dt$

Now,

$\int te^t\,dt=e^t(t-1)+C$

Therefore,

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

Use the initial condition $(1,0)$

At $y=0$, we have $\tan^{-1}0=0$ and $x=1$.

So,

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

Hence the particular solution is

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

Find the abscissa when ordinate is $\tan(1)$

Ordinate means $y$-coordinate, so let

$y=\tan(1)$

Then,

$\tan^{-1}y=1$

Substitute into the solution:

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

Match with the options

$\boxed{\frac{2}{e}}$

This corresponds to Option B.