Question

if Asec@+Btan@+C=0, & psec@+Qtan@+R=0, prove that (br-qc) wholesq. - (aq-bp)wholesq. = (pc-ar) wholesq.

Answer 1

Asecθ+Btanθ+C=0
or, C=-(Asecθ+Btanθ)
Psecθ+Qtanθ+R=0
or, 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]