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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:
$$ \tan(\phi) = \tan\left( \tan^{-1}(m_1) - \tan^{-1}(m_2) \right) \\ =\frac{m_1-m_2}{1+m_1m_2} \qquad \square $$
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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:
$$ RHS = \frac{1}{2}\left(\frac{1}{r} - r\right) = \frac{1}{2}\left(\frac{\cos{\phi}}{1-\sin(\phi)} - \frac{1-\sin(\phi)}{\cos{\phi}}\right) \\ = \frac{1}{2}\left(\frac{ \cos (\phi)^2 - (1-\sin(\phi))^2}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{1}{2}\left(\frac{ 1 - \sin(\phi)^2 - \left(1-2\sin(\phi)+\sin(\phi)^2\right)}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{1}{2}\left(\frac{ 2\sin(\phi) - 2\sin(\phi)^2}{\cos(\phi)(1-\sin(\phi))}\right) \\ = \frac{\sin(\phi)}{\cos(\phi)} = LHS \quad \square $$