# Prove this ?

Aug 22, 2017

see below

#### Explanation:

$l \tan \theta + m \sec \theta = n , l ' \tan \theta - m ' \sec \theta = n '$

Prove

$\implies {\left(\frac{n l ' - \ln '}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{n m ' + m n '}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\frac{\left(l \tan \theta + m \sec \theta\right) l ' - l \left(l ' \tan \theta - m ' \sec \theta\right)}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{\left(l \tan \theta + m \sec \theta\right) m ' + m \left(l ' \tan \theta - m ' \sec \theta\right)}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\frac{l l ' \tan \theta + m l ' \sec \theta - l l ' \tan \theta + l m ' \sec \theta}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{l m ' \tan \theta + m m ' \sec \theta + m l ' \tan \theta - m m ' \sec \theta}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\frac{\cancel{l l ' \tan \theta} + m l ' \sec \theta - \cancel{l l ' \tan \theta} + l m ' \sec \theta}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{l m ' \tan \theta + \cancel{m m ' \sec \theta} + m l ' \tan \theta - \cancel{m m ' \sec \theta}}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\frac{m l ' \sec \theta + l m ' \sec \theta}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{l m ' \tan \theta + m l ' \tan \theta}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\frac{\sec \theta \left(m l ' + l m '\right)}{m l ' + l m '}\right)}^{2} = 1 + {\left(\frac{\tan \theta \left(l m ' + m l '\right)}{l m ' + m l '}\right)}^{2}$

$\implies {\left(\sec \theta \frac{\cancel{\left(m l ' + l m '\right)}}{\cancel{m l ' + l m '}}\right)}^{2} = 1 + {\left(\tan \theta \frac{\cancel{\left(l m ' + m l '\right)}}{\cancel{l m ' + m l '}}\right)}^{2}$

$\implies {\left(\sec \theta\right)}^{2} = 1 + {\left(\tan \theta\right)}^{2}$

$\implies {\sec}^{2} \theta = 1 + {\tan}^{2} \theta$ , but $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

$\implies {\sec}^{2} \theta = {\sec}^{2} \theta$

$\therefore L H S = R H S$

Aug 22, 2017

Please refer to a Proof given in the Explanation Section.

#### Explanation:

We rewrite these eqns. as,

$l \tan \theta + m \sec \theta - n = 0 , \ldots \ldots \ldots \ldots \ldots . . \left(1\right) ,$ and,

$l ' \tan \theta - m ' \sec \theta - n ' = 0 , \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right) .$

Using Kramer's Method to solve these eqns. for

$\tan \theta , \mathmr{and} , \sec \theta .$ we have,

$\tan \frac{\theta}{|} \left(m , - n\right) , \left(- m ' , - n '\right) | = - \sec \frac{\theta}{|} \left(l , - n\right) , \left(l ' , - n '\right) | = \frac{1}{|} \left(l , m\right) , \left(l ' , - m '\right) | ,$

$\therefore \tan \theta = \frac{- \left(m n ' + m ' n\right)}{- \left(l m ' + l ' m\right)} , - \sec \theta = \frac{l ' n - l n '}{-} \left(l m ' + l ' m\right) ,$

Since, ${\sec}^{2} \theta = 1 + {\tan}^{2} \theta ,$ we get,

${\left\{\frac{l ' n - l n '}{l m ' + l ' m}\right\}}^{2} = 1 + {\left\{\frac{m n ' + m ' n}{l m ' + l ' m}\right\}}^{2} ,$ as desired!

Enjoy Maths.!