Question

If A secthita + B tanthita + C =0. and P secthita and Q tanthita +R = 0. prove that (BR-QC) 2 - (PC-AR) 2 = (AQ- BP)2

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]

Answer 2

Answer:

by cross multiplication solve the questions.

this is very easy method.

sec theta /br - qc = c/aq -bp =c/br-qc and solve them.