Question

Let $L_1$and $L_2$denote two nonvertical intersecting lines, and let u denote

the acute angle between $L_1$and $L_2$(see the figure). Show that

$\tan \theta =\frac{m_2-m_1}{1-m_1{m}_2}$

where $m_{1}and{m}_2$ are the slopes of$L_{1}and{L}_2$, respectively.

Short answer

Equation $\tan \theta =\frac{m_2-m_1}{1-m_1{m}_2}$can be obtained by using the difference formula for tangent function as

$$\begin{gathered} \tan \left(\theta \right)=\tan ({\theta }_2-{\theta }_1) \\ \tan \left(\theta \right)=\frac{\tan \left({\theta }_2\right)-\tan \left({\theta }_1\right)}{1-\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)} \\ \tan \theta =\frac{m_2-m_1}{1-m_1{m}_2} \end{gathered}$$

Step-by-step solution

Step 1. Given data

The given figure is

The equation that needs to prove is$\tan \theta =\frac{m_2-m_1}{1-m_1{m}_2}$

Step 2. Formation of trigonometric function

As from figure

$$\begin{gathered} \theta ={\theta }_2-{\theta }_1 \\ \tan \theta =\tan ({\theta }_2-{\theta }_1) \end{gathered}$$

Use difference formula for the tangent function

$$\begin{gathered} \tan ({\theta }_2-{\theta }_1)=\frac{\tan \left({\theta }_2\right)-\tan \left({\theta }_1\right)}{1-\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)} \\ So\tan \theta =\frac{\tan \left({\theta }_2\right)-\tan \left({\theta }_1\right)}{1-\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)} \end{gathered}$$

Step 3. Proof

Substitute $m_1=\tan {\theta }_{1}and{m}_2=\tan {\theta }_2$

$$\begin{gathered} \tan \theta =\frac{\tan \left({\theta }_2\right)-\tan \left({\theta }_1\right)}{1-\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)} \\ \tan \theta =\frac{m_2-m_1}{1-m_1{m}_2} \end{gathered}$$