With oblique coordinates find the tangent of the angle between the straight lines `y = mx + c and my + x = d`.

Answer

To find the tangent of the angle between the straight lines given in oblique coordinates, we will follow these steps:

### Step 1: Identify the equations of the lines
The equations of the lines are:
1. \( y = mx + c \) (Line 1)
2. \( my + x = d \) (Line 2)

### Step 2: Rewrite the second line in slope-intercept form
To find the slope of the second line, we can rearrange the equation \( my + x = d \):
\[ my = d - x \implies y = \frac{d - x}{m} \implies y = -\frac{1}{m}x + \frac{d}{m} \]
Thus, the slope \( m_2 \) of Line 2 is \( -\frac{1}{m} \).

### Step 3: Find the tangent of the angles with respect to the oblique coordinate system
Using the formula for the tangent of the angle with the axis in an oblique coordinate system:
\[ \tan \theta = \frac{m \sin \omega}{1 + m \cos \omega} \]
For Line 1, the slope \( m_1 = m \):
\[ \tan \theta_1 = \frac{m \sin \omega}{1 + m \cos \omega} \]

For Line 2, the slope \( m_2 = -\frac{1}{m} \):
\[ \tan \theta_2 = \frac{-\frac{1}{m} \sin \omega}{1 - \frac{1}{m} \cos \omega} = \frac{-\sin \omega}{m - \cos \omega} \]

### Step 4: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \( \alpha \) between two lines is given by:
\[ \tan \alpha = \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} \]

### Step 5: Substitute the values of \( \tan \theta_1 \) and \( \tan \theta_2 \)
Substituting the values we calculated:
\[ \tan \alpha = \frac{\frac{m \sin \omega}{1 + m \cos \omega} - \left( \frac{-\sin \omega}{m - \cos \omega} \right)}{1 + \left( \frac{m \sin \omega}{1 + m \cos \omega} \right) \left( \frac{-\sin \omega}{m - \cos \omega} \right)} \]

### Step 6: Simplify the expression
After simplification, the numerator becomes:
\[ \frac{m \sin \omega (m - \cos \omega) + \sin \omega (1 + m \cos \omega)}{(1 + m \cos \omega)(m - \cos \omega)} \]
The denominator simplifies to:
\[ 1 - \frac{m \sin^2 \omega}{(1 + m \cos \omega)(m - \cos \omega)} \]

### Final Result
After simplifying the entire expression, we find:
\[ \tan \alpha = \frac{m^2 + 1}{m^2 - 1} \cdot \frac{\sin \omega}{\cos \omega} \]