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In the textbook it states this formula but doesn't go into how it was derived. My geometric intuition is to somehow project the tangent lines out at the point of intersection thus creating a right triangle, but I can't seem to figure out how they get the formula.

The angle between two curves is the angle between their tangent lines at the point of intersection. If the slopes are $m_1$ and $m_2$, then the angle of intersection $\alpha$ can be obtained from the formula $$\tan\alpha=\bigg|\dfrac{m_2-m_1}{1+m_1m_2}\bigg|$$

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    $\begingroup$ As a hint, the slope of a line $m$ is the tangent of the angle that the line makes with the X-axis. So if $m_1=tan(A)$ and $m_2=tan(B)$, what would $|tan(A-B)|$ be? $\endgroup$ – user1952500 Jun 18 '17 at 2:31
  • $\begingroup$ very nice comment! $\endgroup$ – john fowles Jun 18 '17 at 21:34
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Hint: remember the trig identity $\,\tan(\alpha_2-\alpha_1) = \cfrac{\tan(\alpha_2)-\tan(\alpha_1)}{1+ \tan(\alpha_2)\tan(\alpha_1)}\,$ with $m_{1,2}=\tan(\alpha_{1,2})\,$.

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Let the slope and inclination of the first tangent line be $m_1$ and $\theta$, while the second be $m_2$ and $\phi$. It is not difficult to see, for the angel $\alpha$ between them, we have $$\alpha=\pm\theta\mp\phi$$ Thus, $$\tan{\alpha}=\tan{\pm\theta\mp\phi}$$ By compound angel formula, it is easy followed, as $\tan\theta=m_1$ and $\tan\phi=m_2$, $$\tan\alpha=\frac{\pm\tan\theta\mp \tan\phi}{1+\tan\theta\tan\phi}=\frac{\pm m_1\mp m_2}{1+m_1m_2}$$ If we are talking about the acute angle between them, that is, if $\alpha$ is acute, then $$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

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