Math, asked by GuruChoudhary11, 1 year ago

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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Answers

Answered by Anonymous
9
Asecθ+Btanθ+C=0
or, C=-(Asec
θ+Btanθ)
Psec
θ+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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