Q.Show that the tangent of an angle between the lines $\dfrac{x}{a}+\dfrac{y}{b}=1$ and $\dfrac{x}{a}-\dfrac{y}{b}=1$ is $\dfrac{2ab}{a^2-b^2}$.

The key idea is Slope Calculation and the formula for the tangent of the angle between two lines.

1. Convert the given line equations into the slope-intercept form ($y = mx + c$) to find their slopes.

   For the first line, $\dfrac{x}{a}+\dfrac{y}{b}=1 \implies \dfrac{y}{b} = -\dfrac{x}{a} + 1 \implies y = -\dfrac{b}{a}x + b$. So, $m_1 = -\dfrac{b}{a}$.

   For the second line, $\dfrac{x}{a}-\dfrac{y}{b}=1 \implies -\dfrac{y}{b} = -\dfrac{x}{a} + 1 \implies y = \dfrac{b}{a}x - b$. So, $m_2 = \dfrac{b}{a}$.

2. Use the formula for the tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

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

3. Substitute the calculated slopes into the formula:

   $m_2 - m_1 = \dfrac{b}{a} - \left(-\dfrac{b}{a}\right) = \dfrac{b}{a} + \dfrac{b}{a} = \dfrac{2b}{a}$ …

The core idea behind finding the angle between two lines is to first determine their individual slopes. The slope of a line tells us its inclination with respect to the positive x-axis. Once we have the slopes of two lines, say $m_1$ and $m_2$, there's a direct formula to calculate the tangent of the angle between them. This formula is derived from the tangent subtraction identity, relating the angles the lines make with the x-axis.

Let's break down the process:

1. Determine the slopes of the given lines.

   The general form of a linear equation is often $Ax + By + C = 0$. To find the slope, it's usually easiest to convert the equation into the slope-intercept form, $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

   • For the first line: $\dfrac{x}{a}+\dfrac{y}{b}=1$

     To isolate $y$, we first move the term involving $x$ to the right side:

     $\dfrac{y}{b} = 1 - \dfrac{x}{a}$

     $\dfrac{y}{b} = -\dfrac{1}{a}x + 1$

     Now, multiply both sides by $b$:

     $y = -\dfrac{b}{a}x + b$

     Comparing this to $y = m_1 x + c_1$, we find the slope of the first line:

     $m_1 = -\dfrac{b}{a}$

   • For the second line: $\dfrac{x}{a}-\dfrac{y}{b}=1$

     Similarly, isolate $y$:

     $-\dfrac{y}{b} = 1 - \dfrac{x}{a}$

     $-\dfrac{y}{b} = -\dfrac{1}{a}x + 1$

     Multiply both sides by $-b$:

     $y = \dfrac{b}{a}x - b$

     Comparing this to $y = m_2 x + c_2$, we find the slope of the second line:

     $m_2 = \dfrac{b}{a}$

2. Recall the formula for the tangent of the angle between two lines.

   If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$, then the tangent of this angle is given by:

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

   The absolute value ensures that we find the acute angle between the lines. If the problem asks for "an angle", it usually implies the acute one, or simply the magnitude of the tangent.

   Watch out

   This formula is valid only if $1 + m_1 m_2 \neq 0$. If $1 + m_1 m_2 = 0$, it means $m_1 m_2 = -1$, which implies the lines are perpendicular. In this case, the angle $\theta = 90^\circ$, and $\tan \theta$ is undefined. This would happen if $a^2 - b^2 = 0$, as we will see in the denominator. The problem implicitly assumes $a^2 \neq b^2$.

3. Substitute the calculated slopes into the formula.

   We have $m_1 = -\dfrac{b}{a}$ and $m_2 = \dfrac{b}{a}$.

   Let's substitute these into the formula:

   $\tan \theta = \left| \frac{\left(-\frac{b}{a}\right) - \left(\frac{b}{a}\right)}{1 + \left(-\frac{b}{a}\right)\left(\frac{b}{a}\right)} \right|$

4. Simplify the expression.

   First, simplify the numerator:

   Numerator $= -\dfrac{b}{a} - \dfrac{b}{a} = -\dfrac{2b}{a}$

   Next, simplify the denominator: …