Question

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

Answer 1

Here we have 2 linear equations in sec and tan. (you can take sec = x and tan = y for simplicity sake).
Then they solved the 2 linear equations using cross multiplication method.
Refer to the link below to understand cross multiplication method to solve linear equations in 2 variables.
You can use any method to solve for tan and sec.
We have,
$asec\theta +btan\theta +c=0$
And $psec\theta +qtan\theta +r=0$
Solving these two equations by the cross-multiplication for $sec\theta$ and $tan\theta$ we get
$\frac{sec\theta }{br-qc}=\frac{tan\theta }{cp-ar}=\frac{1}{aq-bp}\Rightarrow sec\theta =\frac{br-cq}{aq-bp}andtan\theta =\frac{cp-ar}{aq-bp}$
$se{c}^2\theta -ta{n}^2\theta =1$
$\Rightarrow {\left(\frac{br-cq}{aq-bp}\right)}^2-{\left(\frac{cp-ar}{aq-bp}\right)}^2=1\iff (br-cq{)}^2-(cp-ar{)}^2=(aq-bp{)}^2$
Then they used the identity $se{c}^2-ta{n}^2=1$
Substitute values of sec and tan to get the required results.