Question

If $asec\theta +btan\theta +c=0$ and $psec\theta +qtan\theta +r=0,$ prove that $(br-qc{)}^2-(pc-ar{)}^2=(aq-bp{)}^2$

Answer 1

We have,

$asec\theta +btan\theta +c=0$

and

$psec\theta +qtan\theta +r=0$

Solving these two equations for $sec\theta$ and $tan\theta$ by the cross multiplication we get,

$\frac{sec\theta }{br-qc}=\frac{tan\theta }{cp-ar}=\frac{1}{aq-bp}\Rightarrow sec\theta =\frac{br-cq}{aq-bp}andtan\theta =\frac{cp-ar}{aq-bp}$

$Now,se{c}^2\theta -ta{n}^2\theta =1$

$\Rightarrow {\left(\frac{br-cq}{aq-bp}\right)}^2-{\left(\frac{cp-ar}{aq-bp}\right)}^2=1\Rightarrow (br-cq{)}^2-(cp-ar{)}^2=(aq-bp{)}^2$