Question

if a secA+b tanA+c =0 and p secA+q tanA+r =0
Prove that: (br-qc)^2-(pc-ar)^2=(aq-bp)^2

Answer 1

Asecθ+Btanθ+C=0or, C=-(Asecθ+Btanθ)Psecθ+Qtanθ+R=0or, R=-(Psecθ+Qtanθ)∴, (BR-QC)²-(PC-AR)²=[B{-(Psecθ+Qtanθ)}-Q{-(Asecθ+Btanθ)}]²-[P{-(Asecθ+Btanθ)}-A{-(Psecθ+Qtanθ)}]²=(-BPsecθ-BQtanθ+AQsecθ+BQtanθ)²-(-APsecθ-BPtanθ+APsecθ+AQtanθ)²=(AQ-BP)²sec²θ-(AQ-BP)²tan²θ=(AQ-BP)²(sec²θ-tan²θ)=(AQ-BP)² (Proved)  [∵, sec²θ-tan²θ=1]
hope this may help u