1. Tardigrade
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  4. If L1 and L2 be two non-vertical lines with slopes m1 and m2 respectively. If α1 and α2 are the inclinations of lines L1 and L2, respectively. Let θ and φ be the adjacent angles between the lines L1 and L2. Then, match the terms of Column I with terms of Column II and choose the correct option from the codes given below. Column I Column II A tan θ 1 (m1-m2/1+m1 m2)(. as .1+m1 m2 ≠ 0) B tan φ 2 (m2-m1/1+m1 m2) is positive,(as 1+m1 m2 ≠ 0 ) C θ will be acute and φ will be obtuse 3 (m2-m1/1+m1 m2) is negative, (as 1+m1 m2 ≠ 0 ) D θ will be obtuse and φ will be acute 4 (m2-m1/1+m1 m2)(. as .1+m1 m2 ≠ 0)

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Q. If $L_1$ and $L_2$ be two non-vertical lines with slopes $m_1$ and $m_2$ respectively. If $\alpha_1$ and $\alpha_2$ are the inclinations of lines $L_1$ and $L_2$, respectively. Let $\theta$ and $\phi$ be the adjacent angles between the lines $L_1$ and $L_2$. Then, match the terms of Column I with terms of Column II and choose the correct option from the codes given below.
Column I Column II
A $\tan \theta$ 1 $\frac{m_1-m_2}{1+m_1 m_2}\left(\right.$ as $\left.1+m_1 m_2 \neq 0\right)$
B $\tan \phi$ 2 $\frac{m_2-m_1}{1+m_1 m_2}$ is positive,(as $1+m_1 m_2 \neq 0$ )
C $\theta$ will be acute and $\phi$ will be obtuse 3 $\frac{m_2-m_1}{1+m_1 m_2}$ is negative, (as $1+m_1 m_2 \neq 0$ )
D $\theta$ will be obtuse and $\phi$ will be acute 4 $\frac{m_2-m_1}{1+m_1 m_2}\left(\right.$ as $\left.1+m_1 m_2 \neq 0\right)$

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Solution:

Let $L_1$ and $L_2$ be two non-vertical lines with slopes $m_1$ and $m_2$, respectively. If $\alpha_1$ and $\alpha_2$ are the inclinations of lines $L_1$ and $L_2$ respectively, then
$m_1=\tan \alpha_1 \text { and } m_2=\tan \alpha_2$
Let $\theta$ and $\phi$ be the adjacent angles between the lines $L_1$ and $L_2$. Then,
$\theta=\alpha_2-\alpha_1$ and $\alpha_1, \alpha_2 \neq 90^{\circ}$ (exterior angle theorem)
A. Therefore, $\tan \theta=\tan \left(\alpha_2-\alpha_1\right)$
$=\frac{\tan \alpha_2-\tan \alpha_1}{1+\tan \alpha_1 \tan \alpha_2}=\frac{m_2-m_1}{1+m_1 m_2}, \text { as } 1+m_1 m_2 \neq 0$
B. Since, $\phi=180^{\circ}-\theta$
$\therefore \tan \phi=\tan \left(180^{\circ}-\theta\right)=-\tan \theta$
$= -\frac{m_2-m_1}{1+m_1 m_2}, \text { as } 1+m_1 m_2 \neq 0$
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Now, here arise two cases.
C. Case I If $\frac{m_2-m_1}{1+m_1 m_2}$ is positive, then $\tan \theta$ will be positive and tan $\phi$ will be negative, which means $\theta$ will be acute and $\phi$ will be obtuse.
D. Case II If $\frac{m_2-m_1}{1+m_1 m_2}$ is negative, then $\tan \theta$ will be negative and $\tan \phi$ will be positive, which means that $\theta$ will be obtuse and $\phi$ will be acute.