If a secA + b tanA + c = 0 and p secA + q tanA + r = 0, then prove that (br–qc)²–(aq–bp)² = (pc–ar)².
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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]
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]
GuruChoudhary11:
Thanks
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