Show that the tangent of an angle between the lines

Solution:

Given

$\frac{x}{a}+\frac{y}{b}=1$$\ldots \ldots 1$

$\frac{x}{a}-\frac{y}{b}=1$$\ldots \ldots .2$

Firstly, we find the slope of the given lines

$\frac{x}{a}+\frac{y}{b}=1$

Above equation can be written as

$\Rightarrow \frac{y}{b}=1-\frac{x}{a}$

On rearranging we get

$\Rightarrow y=b-\frac{b}{a} x$

$\Rightarrow y=\left(-\frac{b}{a}\right) x+b$

Since, the above equation is in $\mathrm{y}=\mathrm{m} \mathrm{x}+\mathrm{b}$ form.

So, Slope of the equation 1 is

$m_{1}=-\frac{b}{a}$

Now, finding the slope of the equation 2

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

The above equation can be written as

$\Rightarrow-\frac{y}{b}=1-\frac{x}{a}$

$\Rightarrow-y=b-\frac{b}{a} x$

On rearranging we get

$\Rightarrow y=\left(\frac{b}{a}\right) x-b$

$\Rightarrow y=\left(\frac{b}{a}\right) x+(-1) b$

Since, the above equation is in $y=m x+b$ form. So, Slope of the eq. (ii) is

$\mathrm{m}_{2}=\frac{\mathrm{b}}{\mathrm{a}}$

Let $\theta$ be the angle between the given two lines.

$\tan \theta=\left|\frac{m_{1}-m_{1}}{1+m_{1} m_{2}}\right|$

Putting the values of $m_{1}$ and $m_{2}$ in above eq., we get

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

On simplifying we get

$\Rightarrow \tan \theta=\left|\frac{-2\left(\frac{b}{a}\right)}{1-\left(\frac{b^{2}}{a^{2}}\right)}\right|$

$\Rightarrow \tan \theta=\left|\frac{-2\left(\frac{b}{a}\right)}{\frac{a^{2}-b^{2}}{a^{2}}}\right|$

$\Rightarrow \tan \theta=\left|\frac{-2 a b}{a^{2}-b^{2}}\right|$

$\Rightarrow \tan \theta=\frac{2 a b}{a^{2}-b^{2}}$

Hence the proof.