If $a\sec \theta + b\tan \theta + c = 0$ and ${\text{p}}\sec \theta + q\tan \theta + r = 0$, prove that

${\left( {br - qc} \right)^2} - {\left( {pc - ar} \right)^2} = {\left( {aq - bp} \right)^2} \\$

Hint: Make use of the cross multiplication technique & trigonometric identities to solve this problem.

Complete step-by-step answer:

$a \sec \theta + b\tan \theta + c = 0.........................(1) \\$

$p \sec \theta + q\tan \theta + r = 0..........................(2) \\$

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

$ \dfrac{{\sec \theta }}{{br - qc}} = \dfrac{{\tan \theta }}{{cp - ar}} = \dfrac{1}{{aq - bp}} \\$

$\Rightarrow \sec \theta = \dfrac{{br - qc}}{{aq - bp}} $ & $\tan \theta = \dfrac{{cp - ar}}{{aq - bp}} \\$

Now, you know that,

$ {\sec^2}\theta - {\tan^2}\theta = 1 \\$

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

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

Hence Proved.

Note: In this particular type of question, solve these equations and then apply trigonometry identity to get your answer. Mistakes should be avoided during cross multiplication.