First, $\DeclareMathOperator{\P}{\mathbb{P}} \P(X Y > 0) = \P(X>0, Y>0 ~\text{or}~ X<0, Y<0)$ which is $2 \P(X>0,Y>0) ~~\text{(by symmetry)} $

Let us evaluate this by integrating a bivariate normal density. The density of $(X,Y)$ is given by $$ n(x,y) = \frac1{2\pi\sqrt{1-\rho^2}}\exp\left(-\frac1{2(1-\rho^2)}\cdot (x^2 + y^2 - 2\rho x y)\right) $$ So the sought-for probability is given by $$ \P(XY>0) = 2\int_0^\infty \int_0^\infty n(x,y) \; dx \; dy $$ That can be evaluated by first changing to polar coordinates with $dx\; dy = r \; dr \; d\theta$, integrating first with respect to $r$, which leaves (I leave out details) $$ \P(XY>0)= \frac1{\pi \sqrt{1-\rho^2}} \int_0^{\pi/2} \frac{1-\rho^2}{1-\rho\sin(2\theta)} \; d\theta $$ this one I evaluated by the help of maple, obtaining $$ \P(XY>0) = \frac12 + \frac1\pi \arctan(\frac{\rho}{\sqrt{1-\rho^2}}) $$ which is not the expression guessed (?) by the OP. But we can rewrite the answer above by using the identity $\arctan x = \arcsin(\frac{x}{\sqrt{x^2+1}})$ (see https://en.wikipedia.org/wiki/Inverse_trigonometric_functions), I leave that for the OP. Below a plot of this probability as a function of the correlation $\rho$:

It makes sense tat the probability goes to zero when the correlation approaches $-1$, since then it becomes impossible that $X$ and $Y$ has the same sign.

First, $\DeclareMathOperator{\P}{\mathbb{P}} \P(X Y > 0) = \P(X>0, Y>0 ~\text{or}~ X<0, Y<0)$ which is $2 \P(X>0,Y>0) ~~\text{(by symmetry)} $

Let us evaluate this by integrating a bivariate normal density. The density of $(X,Y)$ is given by $$ n(x,y) = \frac1{2\pi\sqrt{1-\rho^2}}\exp\left(-\frac1{2(1-\rho^2)}\cdot (x^2 + y^2 - 2\rho x y)\right) $$ So the sought-for probability is given by $$ \P(XY>0) = 2\int_0^\infty \int_0^\infty n(x,y) \; dx \; dy $$ That can be evaluated by first changing to polar coordinates with $dx\; dy = r \; dr \; d\theta$, integrating first with respect to $r$, which leaves (I leave out details) $$ \P(XY>0)= \frac1{\pi \sqrt{1-\rho^2}} \int_0^{\pi/2} \frac{1-\rho^2}{1-\rho\sin(2\theta)} \; d\theta $$ this one I evaluated by the help of maple, obtaining $$ \P(XY>0) = \frac12 + \frac1\pi \arctan(\frac{\rho}{\sqrt{1-\rho^2}}) $$ which is not the expression guessed (?) by the OP. But we can rewrite the answer above by using the identity $\arctan x = \arcsin(\frac{x}{\sqrt{x^2+1}})$ (see https://en.wikipedia.org/wiki/Inverse_trigonometric_functions), I leave that for the OP. Below a plot of this probability as a function of the correlation $\rho$:

It makes sense that the probability goes to zero when the correlation approaches $-1$, since then it becomes impossible that $X$ and $Y$ has the same sign.