$X$ and $Y$ have a bivariate normal distribution with $\rho$ as covariance. $X$ and $Y$ are standard normal variables.
I showed that $X$ and $Z= \dfrac{Y-\rho X}{\sqrt{1-\rho^2}}$ are independent standard normal variables.
Using this I need to show that
$$P(X >0,Y>0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).$$
Let $\left( V, W \right)$ be i.i.d standard normal, let $U$ be uniform on $\left[ - \pi, \pi \right)$ independent of $R = \sqrt{V^2 + W^2}$ and let $\varphi = \arcsin \rho$. Then the following vectors have the same distributions (think about the Box-Muller transformation) \begin{eqnarray*} \left( X, Y \right) & \overset{d}{=} & \sqrt{2} \left( V \cos \varphi + W \sin \varphi, W \right)\\ \left( V, W \right) & \overset{d}{=} & R \left( \cos U, \sin U \right) \end{eqnarray*} Implying \begin{eqnarray*} P \left[ X > 0, Y > 0 \right] & = & P \left[ \cos U \cos \varphi + \sin U \sin \varphi > 0, \sin U > 0 \right]\\ & = & P \left[ U \in \left( \varphi - \frac{\pi}{2}, \varphi + \frac{\pi}{2} \right) \cap \left( 0, \pi \right) \right]\\ & = & \frac{\varphi}{2 \pi} + \frac{1}{4} \end{eqnarray*}
Here is an alternative proof:
Let $\phi$ be the density of the standard normal distribution \begin{eqnarray*} P \left[ X > 0, Y > 0 \right] & = & P \left[ X < 0, Y < 0 \right]\\ & = & \int_{- \infty}^0 \phi \left( z \right) \int_{- \infty}^0 \frac{1}{\sqrt{1 - \rho^2}} \phi \left( \frac{x - \rho z}{\sqrt{1 - \rho^2}} \right) \mathrm{d} x \mathrm{d} z\\ & = & \int_{- \infty}^0 \phi \left( z \right) \int_{- \infty}^{- \frac{\rho z}{\sqrt{1 - \rho^2}}} \phi \left( x \right) \mathrm{d} x \mathrm{d} z \end{eqnarray*} Let's call the above integral $h \left( \rho \right)$, then after some simplifications $$\frac{\partial h \left( \rho \right)}{\partial \rho} = \frac{1}{2 \pi \sqrt{1 - \rho^2}} $$ By integrating back (or considering the problem as a first-order ordinary differential equation), $h \left( \rho \right) = \frac{1}{2 \pi} \arcsin \rho + K$ where $K$ is some constant. By the special case of independence, $h \left( 0 \right) = \frac{1}{4}$, you get the final solution $$ P \left[ X > 0, Y > 0 \right] = \frac{1}{2 \pi} \arcsin \rho + \frac{1}{4}$$
Using the OP's hint: The event $\{X>0,Y>0\}$ is the same as the event $\{X>0,Z>\frac{-\rho}{\sqrt{1-\rho^2}}X\}$, where now $X$ and $Z$ are independent standard normal variables. Writing $a:=\frac{-\rho}{\sqrt{1-\rho^2}}$ for brevity, the desired probability is expressible as a double integral involving the joint density of $(X,Z)$: $$ P(X>0,Y>0)=P(X>0,Z>aX)=\int_{x=0}^\infty\int_{z=ax}^\infty\frac1{\sqrt{2\pi}}e^{-x^2/2}\frac1{\sqrt{2\pi}}e^{-z^2/2}\,dz\,dx. $$ Switching to polar coordinates, this equals $$ \begin{align} \int_{\theta=\arctan(a)}^{\pi/2}\int_{r=0}^\infty\frac1{2\pi}e^{-r^2/2}\,rdrd\theta&=\int_{\theta=\arctan(a)}^{\pi/2}\frac1{2\pi}\,d\theta\\ &=\frac1{2\pi}\left(\frac\pi2-\arctan a\right)\\ &=\frac14+\frac1{2\pi}\arctan\frac\rho{\sqrt{1-\rho^2}}; \end{align} $$ for the last equality we substitute $a:=\frac{-\rho}{\sqrt{1-\rho^2}}$ and use the fact that the arctan function is odd. To finish off, remember that if $\theta\in[-\pi/2,\pi/2]$ then $$\theta=\arctan\frac{\rho}{\sqrt{1-\rho^2}}\ \Longleftrightarrow\ \tan\theta=\frac{\rho}{\sqrt{1-\rho^2}}\ \Longleftrightarrow\ \sin\theta=\rho\ \Longleftrightarrow\ \theta=\arcsin \rho.$$
