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If two lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ make an angle $\alpha$ on their intersection. Prove that $$\cos\alpha = \frac{a_1a_2+b_1b_2}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$

I have seen a question with $\sin \alpha$ instead, but the answer uses the $\tan \alpha$, if the easier way of proving this is using $$m_1=-\frac{a_1}{b_1},m_2=-\frac{a_2}{b_2}$$

$$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

I would like to have the logic/demonstration of it too, as I can't see it, I am trying to prove this using geometry and trigonometry, not vectors

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Using values as you have mentioned, $$m_1=-\frac{a_1}{b_1},m_2=-\frac{a_2}{b_2}$$

$$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ $$\tan\alpha=\left|\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right|$$

$$\sec^2\alpha = 1 + tan^2 \alpha = \frac{(a_1^2+b_1^2)(a_2^2+b_2^2)}{(a_1a_2+b_1b_2)^2}$$

$$\cos\alpha = \frac{1}{\sec\alpha} = \frac{a_1a_2+b_1b_2}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$

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  • $\begingroup$ What I wanted to know is why is that the tangent of alpha? $\endgroup$ – Juju9708 Aug 30 '20 at 5:56
  • $\begingroup$ This is because slope of a line $m$ is $tan\theta$. For a more comprehensive explanation, you can look at the derivation here. byjus.com/maths/angle-between-two-lines $\endgroup$ – gemspark Aug 30 '20 at 6:01
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    $\begingroup$ @Juju9704 You can show this either by drawing a bunch of perpendicular lines, so that every angle of interest lies in some triangle with a right angle. But you can also appeal to the formula $\tan (a+b) = \frac{\tan a + \tan b}{1-\tan a \tan b}$. $\endgroup$ – Trebor Aug 30 '20 at 7:59

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