Question

hey mate please help me

Answer 1

Answer:

Step-by-step explanation:

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]

Answer 2

$\textbf{Answer is in Attachment !}$