Let $X=\phi(x,y)$, $Y=\psi(x,y)$ define a transformation of the $xy$-plane to $XY$-plane. Suppose further, $\phi_x=\psi_y$ and $\phi_y=-\psi_x$. Then prove that the angle between the curves $F(x,y)=0$ and $G(x,y)=0$ in the $xy$-plane is equal to the angle between the curves $F_1(X,Y)=0$ and $G_1(X,Y)=0$ in the $XY$-plane, where the transformation maps $F(x,y)$ and $G(x,y)$ to $F_1(X,Y)$ and $G_1(X,Y)$ respectively.
I have no idea to solve this. I know that the angle between the curves $F(x,y)=0$ and $G(x,y)=0$ in $xy$-plane is $$\tan^{-1}{\frac{F_xG_y-F_yG_x}{F_xG_x+F_yG_y}}$$ (as slope of the tangents are $m_1=-\frac{F_x}{F_y}$ and $m_2=-\frac{G_x}{G_y}$)
Similarly the angle between the curves $F_1(X,Y)=0$ and $G_1(X,Y)=0$ in $XY$-plane is $$\tan^{-1}{\frac{F_1{_X}G_1{_Y}-F_1{_Y}G_1{_X}}{F_1{_X}G_1{_X}+F_1{_Y}G_1{_Y}}}$$
How to prove that they are same using the transformation proveded in the question.
Hints: $$\frac {\partial {F(x,y)}}{\partial x}$$ $$=\frac {\partial {F_1(X,Y)}}{\partial X}\frac {\partial X}{\partial x}+\frac {\partial {F_1(X,Y)}}{\partial Y}\frac {\partial Y}{\partial x}$$
$$=\frac {\partial {F_1(X,Y)}}{\partial X}\phi_x+\frac {\partial {F_1(X,Y)}}{\partial Y}\psi_x$$