Relationship between correlation coefficient and the angle between two regression lines From: https://en.wikipedia.org/wiki/Pearson_correlation_coefficient


Geometric interpretation Regression lines for $y = g_X(x)$ [red] and $x = g_Y(y)$ [blue] Regression lines for $y = g_X(x)$ [red] and $x = g_Y(y)$ [blue]
For uncentered data, there is a relation between the correlation
coefficient and the angle $\varphi$ between the two regression lines, $y = g_X(x)$ and $x = gY(y)$, obtained by regressing $y$ on $x$ and $x$ on $y$
respectively. (Here, $\varphi$ is measured counterclockwise within the first
quadrant formed around the lines' intersection point if $r > 0$, or
counterclockwise from the fourth to the second quadrant if $r < 0$.) One
can show[18] that if the standard deviations are equal, then $r = \sec \varphi$
− tan $\varphi$, where sec and tan are trigonometric functions.

How can we prove this relationship mentioned above: $r = \sec \varphi − \tan \varphi$?

Working based on the listed reference ([18]).
The equations of the two regression lines are
$$
x= a_{xy} + b_{xy} y  \\
y= a_{yx} + b_{yx} x
$$
If we graph these, then they will have gradients of $b_{yx}$ and $\frac{1}{b_{xy}}$, respectively.
The relationship between these regression coefficients, the standard deviations ${s_x}, {s_y}$ on $x$ and $y$, and the correlation $r$ is:
$$
b_{xy} = r\frac{s_x}{s_y},\quad \text{and} \quad b_{yx} = r\frac{s_y}{s_x}
$$
So, in the case that $s_x=s_y$, the two regression lines have gradients of $r$ and $1/r$.
In the following, I will assume that $0<r<1$. In the case that $r=1$, the two lines are the same line, and so the angle is zero; and in the case that $r=0$, the two lines are orthogonal. You can show that each of these cases satisfy the identity that you care about with substitution. For the case that $r<0$, the same idea of the proof works, you just need to flip the entire picture upside down in your mind.
Lemma 1: the angle between two lines with gradients $m_1$ and $m_2$ is given by the equation
$$\tan (\phi) = \frac{m_1 - m_2}{1+m_1m_2}$$
Proof: The line with gradient $m_1$ makes an angle of $\theta_1 = \tan^{-1}(m_1)$ with the $x$-axis. The line with gradient $m_2$ makes an angle of $\theta_2 = \tan^{-1}(m_2)$ with the $x$-axis. So the angle between the lines is:
$$
\phi = \theta_1 - \theta_2 = \tan^{-1}(m_1) - \tan^{-1}(m_2) 
$$
Then we can use the compound angle formula for tan:
$$
\tan(\phi) = \tan\left( \tan^{-1}(m_1) - \tan^{-1}(m_2)  \right) \\
=\frac{m_1-m_2}{1+m_1m_2} \qquad \square
$$
--
So, applying this to our equation, we learn that
$$ 
\tan(\phi) = \frac{1}{2}\left(\frac{1}{r} - r\right)
$$
We can show that this equation is satisfied by $r=\sec(\phi)-\tan(\phi)$ by substituting this equation into the RHS, and showing that it reduces to the LHS.
Let $r=\frac{1-\sin(\phi)}{\cos{\phi}}$. Then:
$$
RHS = \frac{1}{2}\left(\frac{1}{r} - r\right) = \frac{1}{2}\left(\frac{\cos{\phi}}{1-\sin(\phi)} - \frac{1-\sin(\phi)}{\cos{\phi}}\right) \\
= \frac{1}{2}\left(\frac{ \cos (\phi)^2 - (1-\sin(\phi))^2}{\cos(\phi)(1-\sin(\phi))}\right) \\
= \frac{1}{2}\left(\frac{ 1 - \sin(\phi)^2 - \left(1-2\sin(\phi)+\sin(\phi)^2\right)}{\cos(\phi)(1-\sin(\phi))}\right) \\
= \frac{1}{2}\left(\frac{ 2\sin(\phi) - 2\sin(\phi)^2}{\cos(\phi)(1-\sin(\phi))}\right) \\
= \frac{\sin(\phi)}{\cos(\phi)} = LHS \quad \square
$$
